The sum of first 16 terms of the AP: 10, 6, 2,... is
a. -320
b. 320
c. -352
d. -400

Solution:

An arithmetic progression (AP) is a sequence where the two consecutive terms have the same common difference. It is obtained by adding the same fixed number to its previous term.

From the question given,

AP is 10, 6, 2,

a = 10

d = - 4.

The formula to find the sum is

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

S₁₆ = 16/2 [2a + (16 - 1)d]

S₁₆ = 8[2 × 10 + 15(-4)]

S₁₆ = 8(20 - 60)

S₁₆ = 8(-40)

S₁₆ = -320.

Therefore, S₁₆ = -320.

✦ Try This: The sum of the third and the seventh terms of an AP is 6 and their product is 8. Find the sum of the first sixteen terms of the AP

☛ Also Check: NCERT Solutions for Class 10 Maths Chapter 5

NCERT Exemplar Class 10 Maths Exercise 5.1 Problem 16

The sum of first 16 terms of the AP: 10, 6, 2,... is, a. -320, b. 320, c. -352, d. -400

Summary:

An arithmetic progression (AP) is a sequence where the two consecutive terms have the same common difference. The sum of the first 16 terms of the AP: 10, 6, 2,... is -320

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