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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