How many terms of the A.P. 8, 15, 22, ldots a | vedclass

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(C) The given $A.P.$ is $8, 15, 22, \ldots$ where the first term $a = 8$ and the common difference $d = 15 - 8 = 7$.Let the number of terms be $n$. Th

Vedclass · Accepted Answer

$20$

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(C) The given $A.P.$ is $8, 15, 22, \ldots$ where the first term $a = 8$ and the common difference $d = 15 - 8 = 7$.Let the number of terms be $n$. The sum of $n$ terms of an $A.P.$ is given by $S_n = \frac{n}{2} [2a + (n - 1)d]$.Given $S_n = 1490$,we have $1490 = \frac{n}{2} [2(8) + (n - 1)7]$.$2980 = n [16 + 7n - 7] = n [7n + 9]$.$7n^2 + 9n - 2980 = 0$.Using the quadratic formula $n = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$,we get $n = \frac{-9 \pm \sqrt{81 - 4(7)(-2980)}}{14} = \frac{-9 \pm \sqrt{81 + 83440}}{14} = \frac{-9 \pm \sqrt{83521}}{14}$.Since $\sqrt{83521} = 289$,$n = \frac{-9 + 289}{14} = \frac{280}{14} = 20$.Thus,$20$ terms are required.

How many terms of the $A.P.$ $8, 15, 22, \ldots$ add up to $1490$?

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