Question

In an ap if S5 +S7 =167 and S10 = 235 then find the AP , where Sn
denotes the sum of its first and n terms

Answer 1

Answer:

Step-by-step explanation:

S5+S7=5/2(2a+4d) +7/2(2a+6d)=167

=5a+10d+7a+21d=167

=12a+31d=167   [let this be eqn 1]

similarly,

from S10=235, we get 2a+9d=47 {let this be equation 2}

by solving eqn 1 and 2, we get a=1 and d=5

therefore the ap is 1,6,11........

Answer 2

First term = a

Common difference = d

By using

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

$= > \: S _{5} = \frac{5}{2}[2a + (5 - 1)d]$

$= > \: \frac{5}{2}[2a + 4d]$

$S_{7} = \frac{7}{2}[2a + (7 - 1)d]$

$= > \: \frac{7}{2}[2a + 6d]$

Given,

$S_{7} - S_{5} = 167$

Therefore,

$\frac{5}{2}[2a + 4d] + \frac{7}{2}[2a + 6d] = 167$

$= > \: 10a + 20d + 14a + 42d = 334$

$= > \: 24a + 62d = 334...(1)$

Given,

$S_{10} = 235$

So, 5(2a + 9d) = 235

=> 2a + 9d = 47 ...(2)

Multiply Eq(2) by 12, we get

24a + 108d = 564...(3)

Subtracting Eq(3) from (1), we get

-48d = -230

Therefore, d = 5

Substing the values of d = 5 in Eq(1), we get

2a + 9(5) = 47 Or 2a = 2

Therefore, a = 1

Then A.P is 1, 6, 11, 16, 21,...