Question

5th term of AP is 21 and 11th term is 39 find first term, common difference and sum of first 5th term of series

Answer 1

$\huge\mathfrak{Solution :}$

Given :-

${a_5}$ = 21

And

${a_1_1}$ = 39

Using formula

${a_n}$ = a + (n - 1)d

We get,

a + 4d = 21 _____(1)

And

a + 10d = 39 _______(2)

Subtract (1) and (2), we get

-6d = -18

$\textbf{d = 3}$

Put this in (1), we get

a + 4(3) = 21

a + 12 = 21

a = 21 - 12

$\textbf{a = 9}$

Now,

${a_5}$ = a + 4d

${a_5}$ = 9 + 4(3)

${a_5}$ = 9 + 12

${a_5}$ = 21

Now,

${S_n}$ = n / 2 (a + ${a_n}$)

${S_5}$ = 5 / 2 (a + ${a_5}$)

${S_5}$ = 5 / 2 (9 + 21)

${S_5}$ = (5 / 2) × 30

${S_5}$ = 5 × 15

${S_5}$ = 75

$\huge\mathbb{Hope \ this \ helps.}$

Answer 2

Hii ☺ !!

5th term = 21

a + 4d = 21

a = ( 21 - 4d ) ----------(1)

And,

11th term = 39

a + 10d = 39 -------(2)

Putting the value of a in equation (2) ,we get

a + 10d = 39

21 - 4d + 10d = 39

6d = 39 - 21

6d = 18

d = 3

Putting the value of d in equation (1) , we get

a = 21 - 4d = 21 - 4 × 3

a = 21 - 12

a = 9

Therefore,

First term ( a ) = 9

Common difference ( d ) = 3

And,

S5 = 5/2 × ( 2a + ( n - 1 ) × d

S5 = 5/2 × [ 2 × 9 + ( 5 - 1 ) × 3 ]

S5 = 5/2 × ( 18 + 12 ) = 5/2 × 30

S5 = 75.