Question

The sum of a series in A.P. is 72,the 1st term being 17 and the common difference -2 the numbers of terms is

Answer 1

Answer:

Step-by-step explanation:

Answer 2

Answer:

The number of terms is 6 or 12.

Step-by-step explanation:

Given

Sum of the series(S)= 72

First-term(a)= 17

Common difference(d)= -2

Let us denote the number of terms as n

We know,

$S=\frac{n}{2}[2a+(n-1)d]$

So by putting the given values equation becomes as,

$72=\frac{n}{2}[2(17)+(n-1)(-2)]$

on multiplying we get,

$72=\frac{n}{2}[34-2n+2]$

Adding the values in the bracket we get,

$72=\frac{n}{2}[36-2n]$

taking 2 common from the bracket,

$72=\frac{n}{2}X2(18-n)$

in the next step after canceling 2 we get,

$72=18n-n^{2}$

taking both right-hand side terms to left-hand side, we get

$n^{2}-18n+72=0$

this is a quadratic equation for which the factor n is

$n=\frac{-b\frac{+}{-}\sqrt[]{b^{2-4ac} } }{2}$, where b is the coefficient of n, a is the coefficient of $n^{2}$ and c is the constant term.

So, putting the values accordingly we get,

$n=\frac{18\frac{+}{-}\sqrt[]{18^{2-4X1X72} } }{2}$

on solving the numbers in the square root we get,

$n=\frac{18\frac{+}{-}\sqrt[]{36} }{2}$

square root of 36 is 6, putting in the equation we get,

$n=\frac{18\frac{+}{-}6 }{2}$

now, $n=\frac{18-6}{2}$ , or $n=\frac{18+6}{2}$

on solving for both the value we have,

     $n=\frac{12}{2}$ , or $n=\frac{24}{2}$

∴ n=6 or n=12.