Question

Find the value of a25 - a15 for the AP: 6, 9, 12, 15, ……​

Answer 1

Answer:

30

Explanation:

Given A. P. :-

• 6, 9, 12, 15

Here,

• first term $\rm(a)$ = 6
• common difference $\rm(d)$ = 3.

We know that,

nth term of an A.P. is given by,

$\implies\rm a_n = a + (n - 1)d$

So 15th term of the AP is:-

$\implies\rm a_{15} = 6 + (15 - 1)3$

$\implies\rm a_{15} = 6 + (14)3$

$\implies\rm a_{15} = 6 + 42$

$\implies\rm a_{15} = 48$

And also, 25th term of the AP:-

$\implies\rm a_{25} = 6 + (25 - 1)3$

$\implies\rm a_{25} = 6 + (24)3$

$\implies\rm a_{25} = 6 + 72$

$\implies\rm a_{25} = 78$

$\rm\therefore \: a_{25} - a_{15} = 78 - 48 \\ \rm \implies a_{25} - a_{15} =\bf 30$

Required answer: 30

__________________________

Answer 2

Given,

AP is 6,9,12,15, ....

$\rm find \: : \: value \: \: of \: \: \: a_{25} \: - a_{15}$

SOLUTION :-

from the given AP, we observe that

First term (a) = 6

Common difference (d) = 9 - 6 = 3

then, as we know that

nth term of an AP is given by

$\ast \: \: \underline{ \boxed{ \color{gold} \bf a_n = a + (n - 1)d}}$

then,

$\tt⇢ \: \:a_ {25} = a + 24d$

$\tt⇢ \: \:6 + 24 \times 3$

$\tt⇢ \: \:6 + 72$

$\tt⇢ \: \:a_ {25} = 78$

and,

$\tt⇢ \: \:a_ {15} = a + 14d$

$\tt⇢ \: \:6 + 14 \times 3$

$\tt⇢ \: \:a_ {15} = 48$

then, on substituting the values

$\tt⇢ \: \: a_ {25} - a_ {15}$

$\tt⇢ \: \:78 - 48$

$\tt⇢ \: \:30$

FINAL ANSWER :-

$\rm \therefore the \: \: value \: \: of \: \: \boxed{ \color{lime}\bf a_ {25} - a_ {15} = 30}$