Question

If Sn denotes the sum of the first n term of an AP prove that S30 = 3(S20 -S10)

Answer 1

Hey there,

Let the first term be a and common difference=d

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

S20=20/2[2a+(20-1)d]=10(2a+19d)

=20a+190d. ...(ii)

S10=10/2[2a+(10-1)d]=5(2a+9d)
=10+45d ..(ii)

S30=30/2[2a+29d)=15(2a+29d)

=>S30=30a+435d ...(iii)

Subtracting eq(ii) from (I) we get

=>S20-S10=10a+145d..(iv)

Now multiplying eq. (iv)by 3 we get,

=>3(S20-S10)=30a+435d..(v)

From eq(iii) and (v)we get

=>S30=3(S20-S10)

Hope it helps

Answer 2

Step-by-step explanation:

Suppose a be first term and d is common difference

$\bf\huge S_{30}  = \frac{30}{2} (2a + 29d)$

= 15(2a + 29d) ................ (1)

$\bf\huge = 3(S_{20}  - S_{10})$

= 3[10 (2a + 19d) - 5(2a + 9d)]

= 3(20a + 190d - 10a - 45d)

= 3[10a + 145d]

= 15[2a + 29d] ...............(2)

LHS = RHS