The sum of how many terms of the AP $ 8,15,22, \ldots $ is $ 1490 ? $

Given:

Given AP is \( 8,15,22, \ldots \).

To do:

We have to find the number of terms whose sum is 1490.

Solution:

We know that,

Sum of $n$ terms of an AP is $S_n=\frac{n}{2}[2a+(n-1)d]$

Here,

First term $a_1=a=8$

Second term $a_2=15$

Common difference $d=a_2-a_1=15-8=7$

Let the number of terms whose sum is 1490 be $n$.

Therefore,

$1490=\frac{n}{2}[2(8)+(n-1)7]$

$2\times1490=n(16+7n-7)$

$2980=n(7n+9)$

$7n^2+9n-2980=0$

$n=\frac{-9 \pm \sqrt{9^2-4\times7\times(-2980)}}{2\times7}$

$=\frac{-9 \pm \sqrt{81+83440}}{14}$

$=\frac{-9 \pm \sqrt{83521}}{14}$

$=\frac{-9 \pm 289}{14}$

$=20$       [$n$ cannot be negative]

The number of terms whose sum is 1490 is 20.

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Updated on: 10-Oct-2022

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