Question

Find the sum of the first 8 terms of an AP whose sum of first 10 terms is 16 and the first 14
terms is 32.

Answer 1

Answer :

The required sum of first 8 terms of the AP is 352/35

Step-by-step explanation :

Given :

• Sum of first 10 terms = 16
• Sum of first 14 terms = 32

To find :

The sum of first 8 terms

Solution :

In an A.P., sum of first n terms is given by,

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

where

a denotes first term

d denotes common difference

Sum of first 10 terms = 16

 $\sf S_{10}=\dfrac{10}{2}[2a+(10-1)d] \\\\ \sf 16=5[2a+9d] \\\\ \sf 16=10a+45d \longrightarrow[1]$

Sum of first 14 terms = 32

 $\sf S_{14}=\dfrac{14}{2}[2a+(14-1)d] \\\\ \sf 32=7[2a+13d] \\\\ \sf 32=14a+91d \longrightarrow [2]$

Multiply the equation [1] by 2,

2(16) = 2(10a + 45d)

 32 = 20a + 90d ➟ [3]

Subtract equation [3] from equation [2],

 32 - 32 = 14a + 91d - (20a + 90d)

  0 = 14a + 91d - 20a - 90d

   0 = d - 6a

   6a = d

Substitute d = 6a in equation [1],

16 = 10a + 45d

16 = 10a + 45(6a)

16 = 10a + 270a

280a = 16

  a = 16/280

  a = 2/35

Sum of first 8 terms :

$\sf S_8=\dfrac{8}{2}[2a+(8-1)d] \\\\ \sf S_8=4[2a+7d] \\\\ \sf S_8=8a+28d$

$\sf S_8=8a+28(6a) \\\\ \sf S_8=8a+168a \\\\ \sf S_8=176a \\\\ \sf S_8=176 \times \dfrac{2}{35} \\\\ \boxed{\sf S_8= \dfrac{352}{35}}$

Therefore, the sum of first 8 terms is 352/35