Question

If the sum of $n$ terms of two arithmetic progressions are in the ratio $7n+1:4n+27$, find the ratio of their ${11}^{th}$ terms.

Answer 1

Given ratio of sum of $n$ terms of two A.P is $(7n+1):(4n+27)$

Let $mth$ term of first A.P is $a+(m-1)d$

Let $mth$ term of second A.P is $a^{\prime }+(m-1)d^{\prime }$

ratio of $mth$ term will be $(a+(m-1\left)d\right):(a^{\prime }+(m-1\left)d^{\prime }\right)$

On multiplying by $2$ we get

$[2a+2(m–1\left)d\right]:[2a^{\prime }+2(m–1\left)d^{\prime }\right]$

$=[2a+(2m–1)–1d]:[2a^{\prime }+(2m–1)–1d^{\prime }]$

$S_{2m-1}:{S^{\prime }}_{2m-1}$

$=\left[7\right(2m–1)+1]:\left[4\right(2m–1)+27]$

$[14m–7+1]:[8m–4+27]$

$=[14m–6]:[8m+23]$

Thus the ratio of $mth$ term of two AP’s is $[14m–6]:[8m+23]$.

Putting $m=11$

$[11\times 14-6]:[8\times 14+23]$

$=148:135$