Question

Let S_n denote the sum first term of arithmetic progression. If S₂₀ = 790 and S₁₀ = 146, then S₁₅ - S₅ is:

Answer 1

The sum of the first n terms of an arithmetic progression (A.P.) is given by:

\(S_n = \frac{n}{2} (2A + (n - 1)D)\)

where A is the first term and D is the common difference.

For n = 20:

\(S_{20} = \frac{20}{2} (2A + 19D) = 790\)

10(2A + 19D) = 790

2A + 19D = 79 ............(i)

For n = 10:

\(S_{10} = \frac{10}{2} (2A + 9D) = 145\)

5(2A + 9D) = 145

2A + 9D = 29 ............(ii)

We can subtract Equation 2 from Equation 1:

(2A + 19D) − (2A + 9D) = 79 − 29

10D = 50

D = 5

Substitute D=5 back into Equation 2:

2A + 9(5) = 29

2A + 45 = 29

2A = 29 − 45

2A = −16

A = −8

For S₁₅:

\(S_{15} = \frac{15}{2}(2A + 14D)\)

\(= \frac{15}{2} (2(-8) + 14(5))\)

\(= \frac{15}{2} (-16 + 70)\)

\(\frac{15}{2} \times 54 = 15 \times 24 = 405\)

For S₅:

\(S_5 = \frac{5}{2}(2A + 4D)\)

\(= \frac{5}{2} (2(-8) + 4(5))\)

\(= \frac{5}{2} (-16 + 20) \)

\(= \frac{5}{2} \times 4 = 5 \times 2 = 10\)

S₁₅ − S₅

= 405 − 10

= 395