Question

what is the sum of the first sixteen terms of the arithmetic sequence 1,5,9,13...?​

Answer 1

Answer:

Given:

• a = 1
• d = 4
• n = 6

To find :

• $S_{16}$

$\sf{\underline{Solution:-}}$

Use the formula of finding sum to n terms in an AP.

★Formula to be used :

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

➣Puttng all the values :

➫$S_{16} = \dfrac{16}{2} [2(1) + (16 - 1)4]$

➫$S_{16} = 8 [2 + 15(4)]$

➫$S_{16} = 8 [2 + 60]$

➫$S_{16} = 8 (62)$

➫$S_{16} = 496$

•°•Sum of first 16 terms ia 496

Answer 2

SOLUTION

TO DETERMINE

The sum of the first sixteen terms of the arithmetic sequence 1 , 5 , 9 , 13 . . . .

EVALUATION

Here the given arithmetic sequence is

1 , 5 , 9 , 13 . . . .

First term = a = 1

Common Difference = d = 5 - 1 = 4

Number of terms = 16

Hence the required sum

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

$\displaystyle \sf{ = \frac{16}{2} \bigg[ (2 \times 1) + 4(16 - 1)\bigg] }$

$\displaystyle \sf{ = 8 \times \bigg[ 2 +60 \bigg] }$

$\displaystyle \sf{ = 8 \times 62 }$

$\displaystyle \sf{ = 496 }$

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