Question

The sum of the first five terms of an AP and the sum of the first seven terms of the same AP is 167 . If the sum of the first ten terms of this AP is 235 . Find the sum of its first twenty terms ? ​

Answer 1

Step-by-step explanation:

Let Sn be the sum of n terms of an AP with first term a and common difference d. Then

Sn=2n(2a+(n−1)d)

putting n=5 in Sn, we get

S5=25(2a+4d)⇒S5=5(a+2d)               ...(1)

putting n=7 in Sn, we get

S7=27(2a+6d)⇒S7=7(a+3d)               ...(2)

Adding (1) and (2), we get

12a+31d=167                                                                    ...(3)

Now,S10=235⇒210(2a+9d)=235

⇒2a+9d=47                                                       ...(4)

On applying 6×(4)−(3), we get

23d=115⇒d=5

Putting d=5 in (3), we get

12a=167−31×5⇒a=1

Therefore, Required sum S20=220[2×1+(20−1)×

5

⇒S20=10[2+95]

⇒S20=970

Answer 2

$\large\bf{\underline\orange{Answer \:↝}}$

We know that, in an A.P,

↪First term = a

↪Common difference = d

↪Number of terms of an AP = n

_______________________

According to the question,

We have,

S5 + S7 = 167

Using the formula for sum of n terms,

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

So, we get,

(5/2) [2a + (5-1)d] + (7/2)[2a + (7-1)d] = 167

5(2a + 4d) + 7(2a + 6d) = 334

10a + 20d + 14a + 42d = 334

24a + 62d = 334

12a + 31d = 167

12a = 167 – 31d …(1)

We have,

S10 = 235

(10/2) [2a + (10-1)d] = 235

5[ 2a + 9d] = 235

2a + 9d = 47

Multiplying L.H.S and R.H.S by 6,

We get,

12a + 54d = 282

From equation (1)

167 – 31d + 54d = 282

23d = 282 – 167

23d = 115

d = 5

Substituting the value of d = 5 in equation (1)

12a = 167 – 31(5)

12a = 167 – 155

12a = 12

a = 1

We know that,

S20 = (n/2) [2a + (20 – 1)d]

= 20/(2[2(1) + 19 (5)])

= 10[ 2 + 95]

= 970

Therefore, the sum of first 20 terms is 970.