Question

How many terms of the A.P 1,5,9....must be taken so that their sum is 2415

Answer 1

Hope this helps you.....✌^_^

Answer 2

Answer:

The required number of terms = 35

Step-by-step explanation:

First term , a = 1

Common Difference , d = 4

Sum = 2415

We need to find number of terms such that their sum is 2415

$S_n=\frac{n}{2}\times (2a+(n-1)\times d)\\\\\implies 2415=\frac{n}{2}\times(2+(n-1)\times 4)\\\\\implies 2415=\frac{n}{2}\times(4n-2)\\\\\implies 2n^2-n-2415=0\\\\\implies (2n+69)(n-35)=0$

Since, number of terms cannot be negative

⇒ n = 35

Hence, The required number of terms = 35