Question

If the 13th term of an A.P is 21, then the sum of the first 25 terms is?
answer it pls its urgent

Answer 1

Step-by-step explanation:

$13th \: term = first \: term+ 12d = 21$

$sum \: \: of \: first \: 25 \: \: terms = \frac{25}{2} (1st \: term + 25th \: term)$

$1st \: term = 21 - 12d$

$25th \: term = 21+ 12d$

hence ,

$sum \: of \: first \: 25 \: terms = \frac{25}{2} (21- 12d + 21 + 12d)$

$= \frac{25}{2} (21 + 21)$

$= \frac{25}{2} (42)$

$= 525$

Answer 2

${\huge{\underline{\underline{\sf{\maltese\; Question:-}}}}}$

Q - If the 13th term of an A.P. is 21 , then the sum of the first 25 terms would be ?

${\huge{\underline{\underline{\sf{\maltese\;Answer\checkmark}}}}}$

★ Given :

• 13 th term ($a_{13}$) = 21

★ To Find :

• Sum of first 25 terms ($S_{n}$)

★ Solution :

Using the Formula ↓

$\rm \: a _{n} = a + (n - 1)d$

so , for 13th term ,

n = 13

$\rm \: 21 = a + 12d$

$\mapsto \rm \: a = 21 - 12d....(i)$

Now ,

Using the Formula ↓

$\rm \: s _{n} = \frac{n}{2} (2a + (n - 1)d)$

Here ,

As we have to find the sum of first 25 terms so ,

n = 25

$\rm \: s_{25} = \frac{25}{2} (2a + (25 - 1)d)$

Put the value of a from eq(i)

$\rm \: s_{25} = \frac{25}{2} (2 \times (21 - 12d) + 24d)$

$\rm \: s _{25} = \frac{25}{2} (42 - \cancel{24d} + \cancel{ 24d})$

$\mapsto \rm \: s_{25} = \frac{25}{ \cancel2} \times \cancel{42}$

$\mapsto \rm \: s _{25} = 25 \times 21$

$\large \mapsto \boxed{\rm \: s_{25} = 525}$

So ,

The sum of first 25 terms would be 525 .