Find the sum of the first 20 terms of the A.P | vedclass

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(C) For the given $A.P.$,the first term $a = 5$,the common difference $d = 13 - 5 = 8$,and $n = 20$.The sum of $n$ terms is given by $S_n = \frac{n}{2

Vedclass · Accepted Answer

$40$

Vedclass · Answer

(C) For the given $A.P.$,the first term $a = 5$,the common difference $d = 13 - 5 = 8$,and $n = 20$.The sum of $n$ terms is given by $S_n = \frac{n}{2}[2a + (n - 1)d]$.$	herefore S_{20} = \frac{20}{2}[2(5) + (20 - 1)8]$$\quad = 10[10 + 152]$$\quad = 10[162] = 1620$.Thus,the sum of the first $20$ terms of the $A.P.$ is $1620$.Now,let the sum of $n$ terms be $S_n = 6440$.$	herefore 6440 = \frac{n}{2}[2(5) + (n - 1)8]$$	herefore 12880 = n[10 + 8n - 8]$$	herefore 12880 = 8n^2 + 2n$$	herefore 8n^2 + 2n - 12880 = 0$Dividing by $2$,we get $4n^2 + n - 6440 = 0$.Solving the quadratic equation: $4n^2 + 161n - 160n - 6440 = 0$$	herefore n(4n + 161) - 40(4n + 161) = 0$$	herefore (4n + 161)(n - 40) = 0$Since $n$ must be a positive integer,$n = 40$ (as $n = -\frac{161}{4}$ is not possible).Thus,the sum of $40$ terms of the $A.P.$ is $6440$.

Find the sum of the first $20$ terms of the $A.P.$ $5, 13, 21, \ldots$ and find the number of terms of this $A.P.$ whose sum is $6440$.

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