Question

please solve Qno.10 which is given in the photo...

Answer 1

let a is the first and d is the common difference of AP
use formula ,
S=n/2 {2a + (n-1) d}

now,
S1 =n/2 {2a +(n-1) d}
S2 =2n/2 {2a +(2n-1) d}
S3 =3n/2 {2a + (3n-1)d}

LHS =S3 =3n/2 {2a + (3n-1)d}

RHS =3(S2 - S1)
=3[2n/2 {2a+(2n-1)d-n/2 {2a +(n-1)d}
=3n/2 [4a +4n-2-2a-n+1]
=3n/2 {2a +(3n-1) d}

LHS = RHS

Answer 2

Let "a" be the 1st term of the AP and "d" be the common difference.

Sum of terms---
S=n/2 [2a + (n-1) d]


S₁ =n/2 [2a +(n-1) d]

S₂ =2n/2 [2a +(2n-1) d]

S₃ =3n/2 [2a + (3n-1)d]

Now, we have to prove both the sides are equal.

LHS
S₃=3n/2 [2a + (3n-1)d]

RHS
 3(S₂ - S₁)
3[2n/2 (2a+(2n-1)d-n/2 {2a +(n-1)d]
3n/2 [4a +4n-2-2a-n+1]
3n/2 [2a +(3n-1) d]

Here we got that LHS = RHS

So, S₃ = 3(S₂ - S₁)

Hope This Helps You!