Question

If the 17th term of an AP exceeds the 13th term of the AP by 15 , then the common difference is

3

3.75

3.5

4.25

​

Answer 1

Answer:

$\sf{The \ common \ difference \ is \ 3.75}$

Given:

• The 17th term of an AP exceeds the 13th term of the AP by 15.

To find:

• The common difference (d).

Solution:

$\sf{According \ to \ the \ given \ condition}$

$\sf{t_{13}+15=t_{17}}$

$\sf{But, \ t_{17} \ is \ 4d \ more \ than \ t_{13}}$

$\sf{\therefore{4d=15}}$

$\sf{\therefore{d=\dfrac{15}{4}}}$

$\sf{\therefore{d=3.75}}$

$\sf\purple{\tt{\therefore{The \ common \ difference \ is \ 3.75}}}$

Answer 2

Given ,

The 17th term of an AP exceeds the 13th term of the AP by 15

Let , the common difference of AP be " d "

We know that , the nth term of an AP is given by

$\large \rm \fbox{ a_{n} = a + (n - 1)d}$

Thus ,

a + (17 - 1)d - (a + (13 - 1)d) = 15

16d - 12d = 15

4d = 15

d = 3.75

$\sf \therefore\underline{The \: common \: difference \: of \: AP \: is \: 3.75 }$