Question

The sum of n terms of an AP, whose first term is 6 and common difference is 40 is equal to the sum of 2n terms of another AP, whose first term is 40 and common difference is 6. Find n.​

Answer 1

Solution:-

For First A.P. :-

a = 6.

d = 40.

& For 2nd A.P :-

a = 40

& d = 6.

A.T.Q.

S(n) = S (2n)

$= > \frac{n}{2} ({2a + (n - 1)d}) = \frac{2n}{2} (2a + (n - 1)d \\ \\ = > n(12 + (n - 1) \times 40) = 2n \: (80 + (n - 1) \times 6) \\ \\ = > n(12 + 40n - 40) = 2n(80 + 6n - 6) \\ \\ = > 40n - 28 = 2(6n + 74) \\ \\ = > 2(20n - 14) = 2(6n + 74) \\ \\ = > 20n - 6n = 74 + 14 \\ \\ = > 14n = 84 \\ \\ = > n = 6$

Hence,

n = 6.

Answer 2

Answer: Sn=S2n

i.e n/2[2a+(n-1)d] = 2n/2[2a+(2n-1)d]

And after substituting the values of a and d, we will get the answer as n=11