Question

Sum of 100 terms of an A. P. whose nth term is (2n+1)is
(a) 10100
(b) (101)
(c) (100)
(d) 10110​

Answer 1

ANSWER:

• Sum of 100 terms = 10200.

GIVEN:

• Number of terms = 100.

• nth term = 2n + 1.

TO FIND:

• Sum of 100 terms.

EXPLANATION:

a = 2(1) + 1 [Value of n = 1 for first term]

a= 2 + 1

a = 3

$\sf t_{100} = 2(100) + 1$

$\sf t_{100} = 200 + 1$

$\sf t_{100}= 201$

$\boxed{\bold{\large{S_n=\dfrac{n}{2}(a +l)}}}$

n = 100

a = 3

l = 201

$\sf S_{100}= \dfrac{100}{2}(3 + 201)$

$\sf S_{100} = 50(204)$

$\sf S_{100} = 10200$

Sum of 100 terms = 10200.

VERIFICATION:

$\boxed{\bold{\large{d = t_2 -t_1}}}$

$\sf t_2 = 2(2) + 1$

$\sf t_2 = 4 + 1$

$\sf t_2 = 5$

$\sf t_1= 3$

$\sf d = 5-3=2$

$\boxed{\bold{\large{S_n=\dfrac{n}{2}(2a +(n-1)d)}}}$

n = 100

a = 3

d = 2

$\sf S_{100}=\dfrac{100}{2}(2(3) +(100-1)2)$

$\sf S_{100}=50(6 +(99)2)$

$\sf S_{100}=50(6 +198)$

$\sf S_{100}=50(204)$

$\sf S_{100}=10200$

HENCE VERIFIED.