## NCERT Solutions for Class 11 Maths Chapter 9 Sequence and Series Exercise 9.2

We are providing NCERT Solutions of Chapter 9 Sequence and Series Exercise 9.2 here that will be useful for every Class 11 students who want to improve their marks in the examination. NCERT Solutions for Class 11 Maths provided here are prepared by Studyrankers are accurate and detailed which will help you in knowing the important points of chapter.

1. Find the sum of odd integers from 1 to 2001.

We have to find the sum 1 + 3 + 5 + ......+2001

Let the n

a + (n –1)d = 2001

a = 1, d = 2

1 + [n – 1]. 2 = 2001 or 2[n –1]

= 2001 – 1 =2000

n – 1 = 1000, n = 1001

Sum = n/2[2a + (n – 1)d]

= 1001/2 [2.1 + (1001 – 1) × 2]

= 1001 × 1001 = 1002001

2. Find the sum of all natural numbers lying between 100 and 1000, which are multiples of 5.

We have to find the sum

105 + 110 + 115 + ........+ 995

Let 995 = n

∴ a + [n – 1]d = 995 or 105 + [n – 1]5 = 995

Dividing by 5,

21 + (n – 1) = 199 or n = 199 – 20 = 179

∴ 105 + 110 + 115 + ........+ 995

= n/2 [2a + (n – 1)d]

= 179/2 [2 × 105 + (179 – 1)5]

= 179/2 [2×105 + 5 × 178] = 98450

3. In an A.P. the first term is 2 and the sum of the first five terms is one fourth of the next five terms. Show that 20

Let d be the common difference of A.P.

∴ Sum of first 5 term = n/2[2a + (n – 1)d]

= 5/2[ 2 × 2 + (5 – 1)d] = 5(2 + 2d)

6

∴ (a + 5d) + (a + 6d) + ………… to 5 terms

Sum of the next 5 terms

= 5/2 [2(a + 5d) + (5 – 1)d]

= 5/2 [2a + 10d + 4d] = 5/2[2a + 14d] = 5(a + 7d)

Sum of first five term

= 1/4 × sum of next 5 terms

5(2 + 2d) = 1/4 × 5 (2 + 7d)

=> 4(2 + 2d) = 2 + 7d ∴ d= -6

∴ T

= 2 + (-114) = - 112

4. How many terms of the A.P. –6, 11/2 -, –5, ..... are needed to give the sum –25?

We have a = –6, d= -(11/2)- (–6)

d = (-11 + 12)/2 = 1/2

Let n be the number of terms so that S

Since, S

=> -25 = n/2[2 . -6 + (n -1) × 1/2]

=> -25 = n/2[-12 + (n -1)/2]

=> -25 = n/2[(-24 + n – 1)/2]

=> -25 = n/2[(n – 25)/2] => –100 = n(n – 25)

=> n

=> (n – 5)(n – 20) = 0

=> n = 5 or n = 20

Both values of n give the required sum.

5. In an A.P., if p

Let a be the first term and d be the common difference of an A.P. then

T

T

Eqn (i) – eqn (ii) gives,

(p – q)d = 1/q – 1/p = (p – q)/pq => d = 1/pq

Putting d = 1/pq in (i), we get

a + (p – 1) 1/pq = 1/q => a + 1/q – 1/pq = 1/q

=> a = 1/pq

Now, S

= pq/2 [2/pq + (pq – 1) 1/pq]

= pq/2 [2/pq + 1 – 1/pq] = pq/2 (1/pq + 1)

S

6. If the sum of a certain number of terms of the A.P. 25, 22, 19, ........ is 116. Find the last term.

a = 25, d = 22 – 25 = –3. let n be the no. of terms

Sum = 116; Sum = n/2[2a + (n – 1)d]

116 = n/2 [50 + (n – 1)(-3)]

Or 232 = n[50 – 3n + 3] = n[53 – 3n]

= -3n

=> 3n

=> (n – 8)(3n – 29) = 0

=> n = 8 or n = 29/3, n ≠ 29/3 ∴ n = 8

∴ Now, T

= 25 – 21

∴ Last term = 4

7. Find the sum of n terms of the A.P., whose k

T

T

T

∴ d = T

a = 6, d = 5

Sum of n term = n/2 [2a + (n – 1)d]

= n/2 [2 × 6 + (n – 1)5]

= n/2 [12 + 5n – 5] = (n(5n + 7))/2

8. If the sum of n terms of an A.P. is (pn + qn

Let S

Putting n = 1, 2

S

S

T

= [(2p + 4q) – (p + q)] = p + 3q

d = T

Common difference of the series is 2q.

9. The sums of n terms of two arithmetic progressions are in the ratio 5n + 4: 9n + 6, Find the ratio of their 18th terms.

Let a

S

S

S

= (2a

S

We have to find

(18

Multiplying the numerator and denominator by 2

(T

Comparing (i) and (iii)

(n – 1)d

Putting n = 35 in (ii)

(2a

=> (2a

(T

10. If the sum of first p terms of an A.P. is equal to the sum of the first q terms then find the sum of the first (p + q) terms.

Let a be the first term and d be the common difference of A.P.

Sum of first p terms = p/2 [2a + (p – 1)d] … (i)

Sum of first q terms = q/2 [2a + (q – 1)d] … (ii)

Equating (i) & (ii)

p/2[2a + (p – 1)d] = q/2[2a + (q – 1)d]

Transposing the term of R.H.S to L.H.S

or 2a(p – 1) + p(p – 1) – q(q – 1)d = 0

=> 2a(p – 1) + [(p

Or 2a(p – 1) + (p – q)[(p + q) – d] = 0

=> (p – q)[2a + (p + q – 1)d] = 0

=> 2a + (p + q – 1)d = 0 …….(iii)

(p ≠ q)

Sum of first (p + q) term

= (p + q)/2 [2a + (p + q – 1)d] = (p + q)/2 × 0 = 0

∴ 2a + (p + q – 1)d = 0 [from (iii)]

11. Sum of the first p, q and r terms of an A.P. are a, b and c respectively.

Prove that

a/p (q – r) + b/q (r – p) + c/r (p – q) = 0

Let A be the first term and D the common difference of A.P.

S

=> p/2[2A + (p – 1)D] = a

=> 1/2[2A + (p – 1)D] = a/p

=> A + 1/2 (p – 1) D = a/p ……(i)

Similarly, A + 1/2 (q – 1)D = b/q ……(ii)

And A + 1/2 (r – 1)D = c/r ………(iii)

Multiplying (i) by (q – r) (ii) by (r – p) (iii) by (p – q) and adding, we get

A(q – r + r – p + p – q) + 1/2 [(p – 1)(q – r) + (q -1)(r – p) + (r – 1)(p – q)]

= a/p (q – r) + b/q (r – p) + c/r (p – q)

=> A(0) + 1/2 (0)D = a/p (q – r) + b/q (r – p) + c/r (p – q)

Hence, a/p (q – r) + b/q (r – p) + c/r (p – q) = 0.

12. The ratio of the sums of m and n terms of an A.P. is m

Let a be the first term and d be the common difference of A.P.

then (m/2 [2a + (m – 1)d]/(n/2[2a + (n – 1)d] = m

=> (2a + (m – 1)d)/(2a + (n – 1)d) = m/n …..(i)

For the ratio of m

T

LH.S of eqn (i) will be identical with (ii) if (m –1) replaces (2m – 2) and (n – 1) replaces (2n – 2)

So, replacing m by 2m – 1 and n by 2n – 1 on both sides of (i), we get

(2a + (2m – 2)d)/(2a + (2n – 2)d) = (2m – 1)/(2n – 1)

13. If the sum of n terms of an A.P. is 3n

Let the sum of n term is denoted by S

∴ S

Put n = 1, 2. T

S

∴ T

∴ Common difference d = T

a = 8, d = 6

m

6m + 2 = 164 => 6m = 164 – 2 = 162

∴ m = 162/6 = 27

14. Insert five numbers between 8 and 26 such that the resulting sequence is an A.P.

Let the five numbers be denoted by T

26 is the 7

Then T

26 = 8 + 6d = 6d = 26 – 8 =18

d = 3

T

T

T

T

T

Required five numbers are 11, 14, 17, 20, 23.

15. If (a

A.M between a and b = (a + b)/2

∴ (a

2a

=> a

=> a

=> (a – b)(a

[a ≠ 0]

=> a

(a/b)

16. Between 1 and 31, m numbers have been inserted in such a way that the resulting sequence is an A.P. and the ratio of 7

Let A

Total number of terms = m + 2

If d is the common difference, then

=> T

A

= 1 + 7 × 30/(m + 1) = (m + 1 + 210)/(m + 1) = (m + 211)/(m + 1)

A

= a + (m – 1)d

= 1 + (m – 1)× 30/(m + 1) = (m + 1 + 30m – 30)/(m + 1)

= (31m – 29)/(m + 1)

Now, A

=> 9m + 1899 = 155m – 145 => 146m = 2044

=> m = 2044/146 = 14 => m = 14

17. A man starts repaying a loan as first installment of ₹ 100. If he increase the installment by ₹ 5 every month, what amount he will pay in the 30

The instalments form an A.P. whose first difference is ₹ 100 and the common difference is ₹ 5.

30

= 100 + 29 × 5

= 100 + 145 = 245

Thus, the 30

18. The difference between any two consecutive interior angles of a polygon is 5°. If the smallest angle is 120°. Find the number of the sides of the polygon.

The angles of a polygon of n sides form an A.P. whose first term is 120° and common difference is 5°.

The sum of interior angles

= n/2[2a + (n -1)d] = n/2 [2 × 120 + (n – 1)5]

= n/2 [240 + 5n – 5] = n/2(235 + 5n)

Also the sum of interior angles = 180 × n – 360

∴ n/2 (235 + 5n) = 180n - 360

Multiplying by 2/5, n(47 + n) = 2(36n – 72)

n(47 + n) = 72n – 144

=> n

=> n

=> (n – 16)(n – 9) = 0

=> n ≠ 16 ∴ n = 9

**Answer**We have to find the sum 1 + 3 + 5 + ......+2001

Let the n

^{ th}term be the last term.a + (n –1)d = 2001

a = 1, d = 2

1 + [n – 1]. 2 = 2001 or 2[n –1]

= 2001 – 1 =2000

n – 1 = 1000, n = 1001

Sum = n/2[2a + (n – 1)d]

= 1001/2 [2.1 + (1001 – 1) × 2]

= 1001 × 1001 = 1002001

2. Find the sum of all natural numbers lying between 100 and 1000, which are multiples of 5.

**Answer**We have to find the sum

105 + 110 + 115 + ........+ 995

Let 995 = n

^{ th}term∴ a + [n – 1]d = 995 or 105 + [n – 1]5 = 995

Dividing by 5,

21 + (n – 1) = 199 or n = 199 – 20 = 179

∴ 105 + 110 + 115 + ........+ 995

= n/2 [2a + (n – 1)d]

= 179/2 [2 × 105 + (179 – 1)5]

= 179/2 [2×105 + 5 × 178] = 98450

3. In an A.P. the first term is 2 and the sum of the first five terms is one fourth of the next five terms. Show that 20

^{th}term is – 112.**Answer**Let d be the common difference of A.P.

∴ Sum of first 5 term = n/2[2a + (n – 1)d]

= 5/2[ 2 × 2 + (5 – 1)d] = 5(2 + 2d)

6

^{th}term = a + (6 – 1)d = a + 5d∴ (a + 5d) + (a + 6d) + ………… to 5 terms

Sum of the next 5 terms

= 5/2 [2(a + 5d) + (5 – 1)d]

= 5/2 [2a + 10d + 4d] = 5/2[2a + 14d] = 5(a + 7d)

Sum of first five term

= 1/4 × sum of next 5 terms

5(2 + 2d) = 1/4 × 5 (2 + 7d)

=> 4(2 + 2d) = 2 + 7d ∴ d= -6

∴ T

_{20}= a + (20 – 1)d = 2 + 19 × (-6)= 2 + (-114) = - 112

4. How many terms of the A.P. –6, 11/2 -, –5, ..... are needed to give the sum –25?

**Answer**We have a = –6, d= -(11/2)- (–6)

d = (-11 + 12)/2 = 1/2

Let n be the number of terms so that S

_{n}= –25Since, S

_{n}= n/2[2a + (n – 1)d]=> -25 = n/2[2 . -6 + (n -1) × 1/2]

=> -25 = n/2[-12 + (n -1)/2]

=> -25 = n/2[(-24 + n – 1)/2]

=> -25 = n/2[(n – 25)/2] => –100 = n(n – 25)

=> n

^{ 2}– 25n + 100 = 0=> (n – 5)(n – 20) = 0

=> n = 5 or n = 20

Both values of n give the required sum.

5. In an A.P., if p

^{th}term is 1/q and q^{th}term is 1/p , prove that the sum of first pq terms is 1/2 (pq +1), where p ≠ q.**Answer**Let a be the first term and d be the common difference of an A.P. then

T

_{p}= 1/q => a + (p – 1)d = 1/q ………(i)T

_{q}= 1/p => a + (q – 1)d = 1/p ……(ii)Eqn (i) – eqn (ii) gives,

(p – q)d = 1/q – 1/p = (p – q)/pq => d = 1/pq

Putting d = 1/pq in (i), we get

a + (p – 1) 1/pq = 1/q => a + 1/q – 1/pq = 1/q

=> a = 1/pq

Now, S

_{pq}= pq/2[2a + (pq – 1)d]= pq/2 [2/pq + (pq – 1) 1/pq]

= pq/2 [2/pq + 1 – 1/pq] = pq/2 (1/pq + 1)

S

_{pq}= 1/2(pq + 1)6. If the sum of a certain number of terms of the A.P. 25, 22, 19, ........ is 116. Find the last term.

**Answer**a = 25, d = 22 – 25 = –3. let n be the no. of terms

Sum = 116; Sum = n/2[2a + (n – 1)d]

116 = n/2 [50 + (n – 1)(-3)]

Or 232 = n[50 – 3n + 3] = n[53 – 3n]

= -3n

^{2}+ 53n=> 3n

^{2}– 53 + 232 = 0=> (n – 8)(3n – 29) = 0

=> n = 8 or n = 29/3, n ≠ 29/3 ∴ n = 8

∴ Now, T

_{8}= a + (8 – 1)d = 25 + 7 . (-3)= 25 – 21

∴ Last term = 4

7. Find the sum of n terms of the A.P., whose k

^{ th}term is 5k + 1.**Answer**T

_{k}=5k = + Putting k = 1, 2T

_{1}= 5 × 1 + 1 = 5 + 1 = 6;T

_{2}= 5 × 2 + 1 = 10 + 1 = 11∴ d = T

_{2}– T_{1}= 11 – 6 = 5a = 6, d = 5

Sum of n term = n/2 [2a + (n – 1)d]

= n/2 [2 × 6 + (n – 1)5]

= n/2 [12 + 5n – 5] = (n(5n + 7))/2

8. If the sum of n terms of an A.P. is (pn + qn

^{2}), where p and q are constants, find the common difference.**Answer**Let S

_{n}be the sum of n term S_{n}= pn + qn^{2}Putting n = 1, 2

S

_{1}= T_{1}= p. 1 + q. 1^{2}= p + q,S

_{2}= T_{1}+ T_{2}= p. 2 + q. 2^{2}= 2p + 4qT

_{2}= (T_{1}+ T_{2}) – T_{1}= S_{2}– S_{1}= [(2p + 4q) – (p + q)] = p + 3q

d = T

_{2}– T_{1}= (p + 3q) – (p + q) = 2qCommon difference of the series is 2q.

9. The sums of n terms of two arithmetic progressions are in the ratio 5n + 4: 9n + 6, Find the ratio of their 18th terms.

**Answer**Let a

_{1}, d_{1}be the first term and common difference of the first A.P. and a_{2}, d_{2}be the first term and common difference of the second A.P. If S_{1}and S_{2}be their sums respectively. ThenS

_{1}= n/2 [2a_{1}+ (n – 1)d_{1}];S

_{2}= n/2[2a_{2}+ (n – 1)d_{2}]S

_{1}/S_{2}= (n/2[2a_{1}+ (n – 1)d_{1}])/(n/2[2a_{2}+ (n – 1)d_{2}])= (2a

_{1}+ (n – 1)d_{2})/(2a_{2}+ (n – 1)d_{2}) ….(i)S

_{1}/S_{2}= (5n + 4)/(9n + 6) ……(ii)We have to find

(18

^{th}term of first A.P.)/(18^{th}term of second A.P) = (a_{1}+ 17d_{1})/(a_{2}+ 17d_{2})Multiplying the numerator and denominator by 2

(T

_{18}of I AP)/(T_{18}of II A.P) = (2a_{1}+ 34d_{1})/(2a_{2}+ 34d_{2}) ……..(iii)Comparing (i) and (iii)

(n – 1)d

_{1}= 34d_{1}∴ n = 34 + 1 = 35Putting n = 35 in (ii)

(2a

_{1}+ (35 – 1)d_{1})/(2a_{2}+ (35 – 1)d_{2}) = (5 × 35 + 4)/(9 × 35 + 6)=> (2a

_{1}+ 34d_{1})/(2a_{2}+ 34d_{2}) = (175 + 4)/(315 + 6) = 179/321(T

_{18}of Ist A.P)/T_{18}of IIInd A.P) = 179/321.10. If the sum of first p terms of an A.P. is equal to the sum of the first q terms then find the sum of the first (p + q) terms.

**Answer**Let a be the first term and d be the common difference of A.P.

Sum of first p terms = p/2 [2a + (p – 1)d] … (i)

Sum of first q terms = q/2 [2a + (q – 1)d] … (ii)

Equating (i) & (ii)

p/2[2a + (p – 1)d] = q/2[2a + (q – 1)d]

Transposing the term of R.H.S to L.H.S

or 2a(p – 1) + p(p – 1) – q(q – 1)d = 0

=> 2a(p – 1) + [(p

^{2}– q^{2}– (p – q)d] = 0Or 2a(p – 1) + (p – q)[(p + q) – d] = 0

=> (p – q)[2a + (p + q – 1)d] = 0

=> 2a + (p + q – 1)d = 0 …….(iii)

(p ≠ q)

Sum of first (p + q) term

= (p + q)/2 [2a + (p + q – 1)d] = (p + q)/2 × 0 = 0

∴ 2a + (p + q – 1)d = 0 [from (iii)]

11. Sum of the first p, q and r terms of an A.P. are a, b and c respectively.

Prove that

a/p (q – r) + b/q (r – p) + c/r (p – q) = 0

**Answer**Let A be the first term and D the common difference of A.P.

S

_{p}= a=> p/2[2A + (p – 1)D] = a

=> 1/2[2A + (p – 1)D] = a/p

=> A + 1/2 (p – 1) D = a/p ……(i)

Similarly, A + 1/2 (q – 1)D = b/q ……(ii)

And A + 1/2 (r – 1)D = c/r ………(iii)

Multiplying (i) by (q – r) (ii) by (r – p) (iii) by (p – q) and adding, we get

A(q – r + r – p + p – q) + 1/2 [(p – 1)(q – r) + (q -1)(r – p) + (r – 1)(p – q)]

= a/p (q – r) + b/q (r – p) + c/r (p – q)

=> A(0) + 1/2 (0)D = a/p (q – r) + b/q (r – p) + c/r (p – q)

Hence, a/p (q – r) + b/q (r – p) + c/r (p – q) = 0.

12. The ratio of the sums of m and n terms of an A.P. is m

^{2}: n^{ 2}. Show that the ratio of m^{th}and n^{ th}term is (2m – 1): (2n – 1).**Answer**Let a be the first term and d be the common difference of A.P.

then (m/2 [2a + (m – 1)d]/(n/2[2a + (n – 1)d] = m

^{2}/n^{2}=> (2a + (m – 1)d)/(2a + (n – 1)d) = m/n …..(i)

For the ratio of m

^{th}term to the n^{ th}term, to find the value ofT

_{m}/T_{n}= (a + (m – 1)d)/(a + (n – 1)d) = (2a + (2m – 2)d)/(2a + (2n – 2)d) …(ii)LH.S of eqn (i) will be identical with (ii) if (m –1) replaces (2m – 2) and (n – 1) replaces (2n – 2)

So, replacing m by 2m – 1 and n by 2n – 1 on both sides of (i), we get

(2a + (2m – 2)d)/(2a + (2n – 2)d) = (2m – 1)/(2n – 1)

13. If the sum of n terms of an A.P. is 3n

^{ 2}+ 5n and its m^{th}term is 164, find the value of m.**Answer**Let the sum of n term is denoted by S

_{n}∴ S

_{n}= 3n^{ 2}+ 5nPut n = 1, 2. T

_{1}= S_{1}= 3.1^{2}+ 5.1 = 3 + 5 = 8;S

_{2}= T_{1}+ T_{2}= 3. 2^{2}+ 5. 2 = 12 + 10 = 22∴ T

_{2}= S_{2}– S_{1}= 22 – 8 = 14∴ Common difference d = T

_{2}– T_{1}= 14 – 8 = 6a = 8, d = 6

m

^{th}term = a + (m – 1)d = 164 => 8 + (m – 1) . 6 = 1646m + 2 = 164 => 6m = 164 – 2 = 162

∴ m = 162/6 = 27

14. Insert five numbers between 8 and 26 such that the resulting sequence is an A.P.

**Answer**Let the five numbers be denoted by T

_{2}, T_{3}, T_{4}, T_{5}, T_{6}and 26 are in A.P.26 is the 7

^{th}term. If d is the common differenceThen T

_{7}= a + (7 – 1) d26 = 8 + 6d = 6d = 26 – 8 =18

d = 3

T

_{2}= a + d = 8 + 3 = 11T

_{3}= a + 2d = 8 + 6 = 14T

_{4}= a + 3d = 8 + 9 = 17T

_{5}= a + 4d = 8 + 12 = 20T

_{6}= a + 5d = 8 + 15 = 23Required five numbers are 11, 14, 17, 20, 23.

15. If (a

^{n}+ b^{n})/(a^{n-1 }+ b^{n – 1}) is the A.M between a and b, then find the value of n.**Answer**A.M between a and b = (a + b)/2

∴ (a

^{n}+ b^{n})/(a^{n – 1}+ b^{n-1}) = (a + b)/22a

^{n}+ 2b^{n}= a^{n}+ ab^{n-1 }+ a^{n-1}b + b^{n}=> a

^{n}a^{n-1}b – ab^{n-1}+ b^{n}= 0=> a

^{n-1}(a – b)-b^{n-1}(a- b) = 0=> (a – b)(a

^{n}– b^{n-1}) + b^{n}= 0[a ≠ 0]

=> a

^{n-1}– b^{n-1}= 0 => a^{n-1}= b^{n-1}(a/b)

^{n-1}= 1 = (a/b)^{0}=> n -1 = 0 => n = 1.16. Between 1 and 31, m numbers have been inserted in such a way that the resulting sequence is an A.P. and the ratio of 7

^{th}and (m – 1)^{th}numbers is 5: 9. Find the value of m.**Answer**Let A

_{1}, A_{2}, .........A_{m}be the m arithmetic means each inserted between 1 and 31.Total number of terms = m + 2

If d is the common difference, then

=> T

_{m+2}= 1 + (m + 2 – 1)d = 31 => (m + 1) d = 30 => d = 30/(m + 1)A

_{7}= T_{8}= a + 7d= 1 + 7 × 30/(m + 1) = (m + 1 + 210)/(m + 1) = (m + 211)/(m + 1)

A

_{m-1}= T_{m}= a + (m – 1)d

= 1 + (m – 1)× 30/(m + 1) = (m + 1 + 30m – 30)/(m + 1)

= (31m – 29)/(m + 1)

Now, A

_{7}/A_{m-1 }= (m + 211)/(31m – 29) = 5/9=> 9m + 1899 = 155m – 145 => 146m = 2044

=> m = 2044/146 = 14 => m = 14

17. A man starts repaying a loan as first installment of ₹ 100. If he increase the installment by ₹ 5 every month, what amount he will pay in the 30

^{th}installment?**Answer**The instalments form an A.P. whose first difference is ₹ 100 and the common difference is ₹ 5.

30

^{th}instalment = T_{30}= a + (30 – 1)d= 100 + 29 × 5

= 100 + 145 = 245

Thus, the 30

^{th}instalment is of ₹ 245.18. The difference between any two consecutive interior angles of a polygon is 5°. If the smallest angle is 120°. Find the number of the sides of the polygon.

**Answer**The angles of a polygon of n sides form an A.P. whose first term is 120° and common difference is 5°.

The sum of interior angles

= n/2[2a + (n -1)d] = n/2 [2 × 120 + (n – 1)5]

= n/2 [240 + 5n – 5] = n/2(235 + 5n)

Also the sum of interior angles = 180 × n – 360

∴ n/2 (235 + 5n) = 180n - 360

Multiplying by 2/5, n(47 + n) = 2(36n – 72)

n(47 + n) = 72n – 144

=> n

^{2}+ (47 – 72)n + 144 = 0=> n

^{2}– 25n + 144 = 0=> (n – 16)(n – 9) = 0

=> n ≠ 16 ∴ n = 9