Question

if the n term of an ap is 4n+ 1 find the sum of the first 15 terms of this AP also find the sum of its
n terms

Answer 1

The given AP is of the form : 4n + 1

n=1=== 4*1+1 = 5

Hence, first term = 5 && the common difference = 4 [coefficient of 'n' is the common difference]

The A.P is 5,9,13,....

To find: Sum of first 15 terms

Sum = n/2 * [2*a + (n-1)*d]...  [n-number of terms, a-first term, d-common difference]

Sum = 15/2 * [2*5 + (15-1)*4] = 15/2 * [10 + 14*4] = 15/2 * 66 = 15*33 = 330+165= 495

Hence, sum of first 15 terms = 495.

Sum of n terms:

Sum = n/2 * [2*a + (n-1)*d]

      = n/2 * [ 2*5 + (n-1)*4]

     = n/2 * [ 10 + 4n - 4]

    = n/2 * [6+4n]

   = 6n/2 + 4n^2/2 = 3n + 2n^2

Hence, sum of 'n' terms = 2n^2 + 3n

Hope it helps