Question

the ratio of the sum of n terms of two AP's is (7n+1):(4n+27).find the ratio of their m th terms.

Answer: let a1 , a2 be the 1st terms and d1 , d2 the common differences of the two given A.P's. then the sums of their n terms are given by

Sn = n/2 {2.a1+(n-1)d1} and Sn' = n/2{2. a2 +(n-1)d2}

Sn/Sn' = n/2{2.a1+(n-1)d1} / n/2{2.a2 + (n-1)d2}

Sn / Sn' = 2.a1+(n-1)d1 / 2.a2 + (n-1)d2

it is given that

Sn / Sn' = 7n+1 / 4n + 27

2.a1 + (n-1)d1 / 2.a2 + (n-1)d2 = 7n+1 / 4n+27 ........................ (i)

To find the ratio of the mth terms of the two given AP's , we replace n by (2m-1) in equation (i)

therefore, 2.a1 + (n-1)d1 / 2.a2 + (n-1)d2 = 7(2m-1) + 1 / 4(2m-1) + 27

a1 + (m-1)d1 / a2 + (m-1)d2 = 14m - 6 / 8m + 23

Hence, the ratio of the mth terms of the two A.P's is (14m - 6) : (8m + 23)

My question is, why it has been assumed (2m-1) in the place of 'n' ? has it been arrived from solving with the help of a formula or is it a mere assumption? if it is simply an assumption, why it should be assumed as (2m-1) ? why not 2m or (m - 1)

Answer 1

beasuse the real formula is. tn-1 .but we wants mth terms so we take tm-1