Question

The sum of 15 terms of AP 4,7,10........
is.

Answer 1

Given :

• A.P = 4 , 7 , 10 , .......

• First term ( a ) = 4

• Common difference ( d ) = 7 - 4 = 3

• Number of terms ( n ) = 15

To Find :

• Sum of 15 terms of given A.P

Solution :

$\large \boxed{ \boxed{ \sf S_n = \frac{n}{2} \bigg[2a + (n - 1)d \bigg]} }\\ \\ \implies \sf S_{15} = \frac{15}{2}\bigg[2 \times 4 + (15 - 1)3 \bigg] \\ \\ \implies \sf S_{15} = \frac{15}{2}\bigg[8 + (14)3 \bigg] \\ \\ \implies \sf S_{15} = \frac{15}{2}\bigg[8 + 42 \bigg] \\ \\ \implies \sf S_{15} = \frac{15}{2} \times 50 \\ \\ \implies \sf S_{15} =15 \times 25 \\ \\ \large \implies \boxed{ \boxed{ \sf S_{15} =375}}$

Answer 2

Answer:

$\bullet\sf{A.P. \:is \:4,7,10}$

$\bullet\sf{First\: term\: (a_1)=\: 4}$

$\bullet\sf{Common \:difference,d \:= \:7-4=3}$

$\bullet\sf{number\:of\:terms,n \:= \:15}$

$\large\sf\implies{To\:find\: -Sum\: of\: terms ,S_n}$

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

$\large\implies\sf {\sf{\boxed{\boxed{\sf{S_n= \frac{n}{2}[2a+(n-1)d }}}}}$

$\large\implies\sf {S_n=\frac{15}{2} [2(4)+(15-1)3]}$

$\large\implies\sf {S_n=\frac{15}{2} [8+(14)3]}$

$\large\implies\sf {S_n= \frac{15}{2} [8+42]}$

$\large\implies\sf {S_n= \frac{15}{2}[50]}$

$\large\implies\sf {S_n= \frac{15}{2} \times \cancel 50 }$

$\large\implies\sf {S_n=15 \times25}$

$\large\implies\sf {S_n= 325}$

$\huge\implies\boxed{\boxed{\underline{\red{\bold{ S_n\:=\:375 }}}}}$