Question

13.The sum of all terms in a finile AP is 90. IT first term is 2 and common difference
is 8 then lind number of terms and last term in this AP​

Answer 1

So the given AP is 2 , 10 , 18 ,.... and it is a finite AP.

Sum of all terms of the given AP is 90.

Sum of n terms of an AP is given by the formula

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

where a is the first term and d is the common difference.

Substitute known values to this formula, we get

$90 = \frac{n}{2}[2*2+(n-1)8]$

$180 = n[4+8n-8]$

$180 = -4n+8n^2$

$2n^2-n-45 = 0$

$2n^2-10n+9n-45 = 0$

$2n(n-5)+9(n-5)=0$

$(2n+9)(n-5) = 0$

$n = \frac{-9}{2}$      OR     $n = 5$

Since number of terms of an AP can't be negative or fractional, we reject

$n = \frac{-9}{2}$.

So the AP has in total 5 terms.

The last term is

$a_5 = a+4d$

$a_5 = 2+4*8$

$a_5 = 34$

So the given AP has 5 terms and its last term is 34