Question

Find the sum to 20 terms of an A.P. whose 4th term is 7 and 7th term is 13.​

Answer 1

S(20) = ?

a(4)=7, a(7)=13.

$a4 = a + (4 - 1)d \\ 7 = a + 3d \: \: \: \: \: \: \: \: \: \: \: \:$

$a7 = a + (7 - 1)d \\ 13 = a + 6d \: \: \: \: \: \: \: \: \: \:$

Subtracting the eq:

13=a+6d

7=a+3d

- - -

6 = 3d

d=6/3

d=2

Substituting d=2 in eq.

7=a+3d

7=a+3(2)

7=a+6

a=1

$s(20) = \frac{n}{2} ( \: 2a + (n - 1)d)$

$s(20) = \frac{20}{2} (2(1) + (20 - 1)(2))$

$s(20) = 10(2 + 19 (2))$

$s(20) = 10(2 + 38)$

$s(20) = 10(40)$

$s(20) = 400$

Sum of the first 20 numbers is 400.

Answer 2

SOLUTION

The fourth term of the AP = 7.

That is , a + 3d = 7 $\longrightarrow (1)$  (where d is the common difference and a is the first term of AP)

Also , given that the 7th term of the AP is 13 .

i.e , a + 6d = 13 $\longrightarrow (2)$

Now, let us subtract Equation (1) from Equation (2) , then

a + 6d = 13 - (a + 3d = 7)

a +  6d = 13

-a - 3d = -7

      3d = 6

∴ Common Difference (d) = 6/3 = 2

Now let us substitute the value of d in Equation (2) to obtain value of 'a'

On Substituting ,

a + 6 × 2 = 13

a + 12 = 13

a = 13 - 12 = 1

∴ The first term (a) = 1

Now , we have to find the sum of 20 terms of the AP

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

$S(20) = \dfrac{20}{2} [ 2 *1 + (20-1) 2]$

S(20) = 10 ( 2 + 19 × 2

S(20) = 10 × 38 + 2

S(20) = 10 × 40 = 400

Sum of 20 Terms of the AP is 400