Consider an A.P.: a_1,a_2, ,a_n;a_10.

Question

Consider an A.P.: ${a}_{1},{a}_{2},\ldots ,{a}_{n};{a}_{1}>0$. If ${a}_{2}-{a}_{1}=\frac{-3}{4},{a}_{n}=\frac{1}{4}{a}_{1}$, and $\sum _{i=1}^{n}{a}_{i}=\frac{525}{2}$, then $\sum _{i=1}^{17}{a}_{i}$ is equal to

Accepted Answer

Given ${a}_{1},{a}_{2},\ldots ,{a}_{n}$ are in A.P; ${a}_{1}>0$

Let $d$ be the common difference of this A.P.

So $d={a}_{2}-{a}_{1}=-\frac{3}{4}$ (given)

It is given that $\sum _{i=1}^{n}{a}_{i}=\frac{525}{2}$ and ${a}_{n}=\frac{1}{4}{a}_{1}$

$	herefore $ ${a}_{1}+{a}_{2}+{a}_{3}+\ldots {a}_{n}=\frac{525}{2}$

Using sum of $n$ terms of A.P ${S}_{n}=\frac{n}{2}({a}_{1}+{a}_{n})$

$\Rightarrow$ $\frac{n}{2}({a}_{1}+\frac{1}{4}{a}_{1})=\frac{525}{2}$

$\Rightarrow$ $5{a}_{1}n=525	imes 4$

${a}_{1}n=420-(1)$

Also using formula for general term

${a}_{n}={a}_{1}+(n-1)d$

$\Rightarrow$$\frac{1}{4}{a}_{1}={a}_{1}+(n-1)(-\frac{3}{4})$

$\Rightarrow$ $(n-1)\frac{3}{4}={a}_{1}-\frac{1}{4}{a}_{1}=\frac{3{a}_{1}}{4}$

$n={a}_{1}+1-(2)$

Substitute in equation (1)

${a}_{1}({a}_{1}+1)=420$

$\Rightarrow$ ${a}_{2}^{1}+{a}_{1}-420=0$

$\Rightarrow$ ${a}_{2}^{1}+21{a}_{1}-20{a}_{1}-420=0$

$\Rightarrow$ ${a}_{1}({a}_{1}+21)-20({a}_{1}+21)=0$

$\Rightarrow$ $({a}_{1}+21)({a}_{1}-20)=0$

Since ${a}_{1}>0$ so ${a}_{1}=20$

To find $\sum _{i=1}^{17}{a}_{i}={a}_{1}+{a}_{2}+\ldots {a}_{17}$

Sum of 17 terms $=\frac{17}{2}[2{a}_{1}+(17-1)d]$

$\begin{gathered} =\frac{17}{2}[2	imes 20+16	imes -\frac{3}{4}] \ =\frac{17}{2}[40-12]=\frac{17}{2}	imes 28 \ =17	imes 14=238 \end{gathered}$

Solutions · 1