Question

the sum of the first n terms of an AP whose first term is 8 and the common difference is 20 is equal to the sum of first 2n terms of another AP whose first term is -30 and the common difference is 8. Find n.​

Answer 1

ANSWER: n=13

EXPLANATION -


GIVEN :
1ST AP- a=8 d=20

2ND AP- a=-30 d=8

Sn=S2n
n/2( 2a + n-1 × d) = 2n/2( 2a + 2n-1 × d)

n/2( 2(8) + n-1 × 20) = 2n/2( 2(-30)+ 2n-1 × 8)

n/2(16+ 20n - 20) = n ( -60 + 16n-8)

n(16+ 20n - 20) = 2n ( -60 + 16n-8)

16n+ 20n²-20n = -120n+32n²-16n

8n+10n²-10n=-60n+16n²-16n

-2n+10n²= 16n²-76n

n+5n²=8n²-38n

n+38n=8n²-5n²

39n= 3n²

39=3n

n=39/3

n=13

hence the value of n is 13

hope this helps

Answer 2

$\large\underline\pink{Given:-}$

• The sum of first n term of an Ap. 1st term is 8 and common difference is 20.
• The sum of first 2nd term of another Ap. 1st term is -30 and common difference is 8.

$\large\underline\pink{To find:-}$

• Fine the n ....?

$\large\underline\pink{Solutions:-}$

• The sum of first n term of an Ap. 1st term is 8 and common difference is 20.

$\: \: \: \: \: \leadsto \: \: S \: \: = \: \: \frac{n}{2} \: {[{2a} \: + \: {(n \: - \: {1})} \: d]}$

$\: \: \: \: \: \leadsto \: \: {S_{1}} \: \: = \: \: \frac{n}{2} \: {[{2} \: {(8)} \: + \: {(n \: - \: {1})} \: {20}]}$

$\: \: \: \: \: \leadsto \: \: {S_{1}} \: \: = \: \: \frac{n}{2} \: {[{16} \: + \: {({20n} \: - \: {20})}]}$

$\: \: \: \: \: \leadsto \: \: {S_{1}} \: \: = \: \: \frac{n}{2} \: {[{20n} \: - \: {4}]}$

$\: \: \: \: \: \leadsto \: \: {S_{1}} \: \: = \: \: \frac{n}{2} \: \times \: {2} \: {[{10n} \: - \: {2}]}$

$\: \: \: \: \: \leadsto \: \: {S_{1}} \: \: = \: \: {n} \: {[{10n} \: - \: {2}]}$

$\: \: \: \: \: \leadsto \: \: {S_{1}} \: \: = \: \: {10n}^{2} \: - \: {2n} \: \: \: \: \: \: ....{(i)}.$

• The sum of first 2nd term of another Ap. 1st term is -30 and common difference is 8.

$\: \: \: \: \: \leadsto \: \: S \: \: = \: \: \frac{n}{2} \: {[{2a} \: + \: {(n \: - \: {1})} \: d]}$

$\: \: \: \: \: \leadsto \: \: {S_{2}} \: \: = \: \: \frac{2n}{2} \: {[{2} \: {(-30)} \: + \: {({2n} \: - \: {1})} \: {8}]}$

$\: \: \: \: \: \leadsto \: \: {S_{2}} \: \: = \: \: {n} \: {[{-60} \: + \: {16n} \: - \: {8}]}$

$\: \: \: \: \: \leadsto \: \: {S_{2}} \: \: = \: \: {n} \: {[{16n} \: - \: {68}]}$

$\: \: \: \: \: \leadsto \: \: {S_{2}} \: \: = \: \: {16n}^{2} \: - \: {68n} \: \: \: \: \: \: ....{(ii)}.$

Now, Solving the Eq. (i) and (ii). we get,

$\: \: \: \: \: \: \: {S_{1}} \: \: = \: \: {S_{1}}$

$\: \: \: \: \: \leadsto \: \: {10n}^{2} \: - \: {2n} \: \: = \: \: {16n}^{2} \: - \: {68n}$

$\: \: \: \: \: \leadsto \: \: {10n}^{2} \: - \: {16n}^{2} \: \: = \: \: {2n} \: - \: {68n}$

$\: \: \: \: \: \leadsto \: \: {-6n}^{2} \: \: = \: \: {66n}$

$\: \: \: \: \: \leadsto \: \: {-6n} \: {(n \: - \: {11})} \: \: = \: \: {0}$

$\: \: \: \: \: \leadsto \: \: n \: - \: {11} \: \: = \: \: {0}$

$\: \: \: \: \: \leadsto \: \: n \: \: = \: \: {11}$