Question

2,7,12,17,......sum of 12 terms of this A. P. is​

Answer 1

$\displaystyle\huge\red{\underline{\underline{Solution}}}$

TO DETERMINE

The sum of 12 terms of the Arithmetic progression

2 , 7 , 12 , 17 ,......

FORMULA TO BE IMPLEMENTED

Sum of first n terms of an arithmetic progression

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

Where First term = a

Common Difference = d

CALCULATION

The given Arithmetic progression is

2 , 7 , 12 , 17 ,......

First term = a = 2

Common Difference = d = 12 - 7 = 5

Sum to be determined for 12 terms

So n = 12

Hence the required sum

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

$= \displaystyle \sf{ \frac{12}{2} \bigg[(2 \times 2) + (12 - 1) \times 5 \bigg]}$

$= \displaystyle \sf{ 6 \times(4 + 55) }$

$= \displaystyle \sf{ 6 \times59 }$

$= \displaystyle \sf{ 354 }$

━━━━━━━━━━━━━━━━

LEARN MORE FROM BRAINLY

The sum of the third and seventh term of an AP is 40 and the sum sixth and 14th terms is 70 .Find the sum of first ten terms

https://brainly.in/question/22811954

Answer 2

Here, first term (a) = 2, common difference (d) = 5 and                                   the number of terms (n) = 12

We have to find, the sum of 12th terms of given A. P. $(S_{12})$.

Solution:

We know that:

The sum of nth terms of given A. P. $(S_{n})$.

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

∴ The sum of 12th terms of given A. P. $(S_{12})$

$S_{12} =\dfrac{12}{2} [2\times 2+(12-1)5]$

     = 6[4 + 11 × 5]

     = 6[4 + 55]

     = 6[59]

     = 354

∴ The sum of 12th terms of given A. P. $(S_{12})$ = 354

Thus, the sum of 12th terms of given A. P. $(S_{12})$ is "equal to 354".