Question

the sum of the first 25 terms of ap whose 2nd term is 9 and 4th term is 21

Answer 1

$\textbf{Formula used:}$

$\text{The n th term of the A.P a, a+d, a+2d, ....... is}$

$\boxed{\bf\,t_n=a+(n-1)d}$

$\text{The sum of n terms of an A.P a, a+d, a+2d,.... is}$

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

$\textbf{Given:}$

$t_2=9\;\text{and}\;t_4=21$

$t_2=9\implies\;a+d=9$...(1)

$t_4=21\implies\;a+3d=21$...(2)

$\text{subtracting (2) from (1)}$

$-2d=12$

$\implies\bf\;d=6$

$\text{Put d=6 in (1), we get}$

$a+6=9$

$\implies\bf\;a=3$

$\text{Now, we find sum of first 25 terms of the A.P}$

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

$S_{25}=\frac{25}{2}[2(3)+(25-1)6]$

$S_{25}=\frac{25}{2}[6+(24)6]$

$S_{25}=\frac{25}{2}[6+144]$

$S_{25}=\frac{25}{2}[150]$

$S_{25}=25[75]$

$\implies\boxed{\bf\,S_{25}=1875}$

$\therefore\textbf{The sum of first 25 terms is 1875}$

$\textbf{Find more:}$

1.The sum of first q terms of an A.P. is 63q – 3q². If its pth term is-60, find the value of p. Also, find the 11th term of this A.P.

https://brainly.in/question/15930752

2.The sum of first m terms of an A.P. is 4 m² - m. If its nth term is 107, find the value of n. Also, find the 21st term of this A.P.

https://brainly.in/question/15930761

3.AP given that the first term (a) = 54, the common difference

(d) = -3 and the nth term (an) = 0, find n and the sum of first n terms (Sn)

of the A.P.​

https://brainly.in/question/15922933

Answer 2

Step-by-step explanation:

Formula used:

\text{The n th term of the A.P a, a+d, a+2d, ....... is}The n th term of the A.P a, a+d, a+2d, ....... is

\boxed{\bf\,t_n=a+(n-1)d} t n

=a+(n−1)d

\text{The sum of n terms of an A.P a, a+d, a+2d,.... is}The sum of n terms of an A.P a, a+d, a+2d,.... is

\boxed{\bf\,S_n=\frac{n}{2}[2a+(n-1)d]} Sn = 2n

[2a+(n−1)d]

\textbf{Given:}Given:

t_2=9\;\text{and}\;t_4=21t 2

=9andt 4

=21

t_2=9\implies\;a+d=9t 2

=9⟹a+d=9 ...(1)

t_4=21\implies\;a+3d=21t 4

=21⟹a+3d=21 ...(2)

\text{subtracting (2) from (1)}subtracting (2) from (1)

-2d=12−2d=12

\implies\bf\;d=6⟹d=6

\text{Put d=6 in (1), we get}Put d=6 in (1), we get

a+6=9a+6=9

\implies\bf\;a=3⟹a=3

\text{Now, we find sum of first 25 terms of the A.P}Now, we find sum of first 25 terms of the A.P

S_n=\frac{n}{2}[2a+(n-1)d]S n = 2n

[2a+(n−1)d]

S_{25}=\frac{25}{2}[2(3)+(25-1)6]S

25

= 225

[2(3)+(25−1)6]

S_{25}=\frac{25}{2}[6+(24)6]S

25

= 225

[6+(24)6]

S_{25}=\frac{25}{2}[6+144]S

25 = 225

[6+144]

S_{25}=\frac{25}{2}[150]S

25 = 225

[150]

S_{25}=25[75]S 25=25[75]

\implies\boxed{\bf\,S_{25}=1875}⟹ S 25

=1875

\therefore\textbf{The sum of first 25 terms is 1875}∴The sum of first 25 terms is 187