Question

If the sum of first 7 terms and 15 terms of an AP are 98 and 390 respectively, then find the sum of
first 10 terms​

Answer 1

S7= 7/2 [a+(n-1)d]

98= 7/2 [a+ 6d]

14= 1/2[a+6d]

28=a+6d

a= 28-6d(1)

S15= 15/2[a+(15-1)d]

390= 15/2[a+14d]

26= 1/2[a+14d)

26×2= a + 14d

putting the value of a

52= 28-6d +14d

52-28= 8d

24= 8d

d= 24/8= 3

therefore

a= 28-6d

= 28- 6×3

= 28-18

= 10

S10= 10/2[a+(n-1)d]

= 5[10+9×3]

= 5× 37

= 185

Answer 2

Answer:

185

Step-by-step explanation:

Given,

Sum of 7 terms in A.P = 98

Sum of 15 terms in A.P = 390

To Find :-

Sum of first 10 terms in A.P.

Formula Used :-

Sum of first 'n' terms in A.P =

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

Where,

n = Number of terms.

a = First term

d = Common Difference

Solution:-

Sum of 7 terms in A.P = 98

$\sf\implies S_7 = \dfrac{7}{2}[2a+(7-1)d]$

$\sf 98 = \dfrac{7}{2}[2a+(6)d]$

$\sf 98\times \dfrac{2}{7} = 2a + 6d$

$\sf 14\times 2 = 2(a+3d)$

$\sf\dfrac{28}{2} = a + 3d$

14 = a + 3d

[Let it be equation (1)]

Sum of 15 terms in A.P = 390

$\sf\implies S_15 = \dfrac{15}{2}[2a + (15-1)d]$

$\sf 390 = \dfrac{15}{2}[2a + (15-1)d]$

$\sf 390\times \dfrac{2}{15}[2a + 14d]$

$\sf 26\times 2 = 2(a + 7d)$

$\sf \dfrac{52}{2} = a + 7d$

26 = a + 7d

[Let it be equation (2)]

equation (2) - equation (1)

26 - 14 = a + 7d - a - 3d

12 = 4d

d = $\sf\dfrac{12}{4}$

d = 3

Common difference = d = 3.

Substituting value of d in equation 1:-

14 = a + 3(3)

14 = a + 9

a = 14 - 9

First term = a = 5.

Sum of first 10 terms :-

n = 10

a = 5

d = 3

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

$\sf S_{10} = \dfrac{10}{2}[2(5)+(10-1)3]$

$\sf S_{10} = 5[10 + 9(3)]$

$\sf S_{10} = 5[10 + 27]$

$\sf S_{10} = 5[37]$

$\sf S_{10} = 185$

Sum of first ten terms in A.P = 185