Question

Show that \(a_1, a_2, . . ., a_n, .....\) form an AP where an is defined as below:

1. \(a_n = 3 + 4n\)
2. \(a_n = 9 – 5n\)

Also find the sum of the first 15 terms in each case.

Answer 1

(i) a_n = 3 + 4n
a₁ = 3 + 4(1) = 7
a₂ = 3 + 4(2) = 3 + 8 = 11
a₃ = 3 + 4(3) = 3 + 12 = 15
a₄ = 3 + 4(4) = 3 + 16 = 19 
It can be observed that
a₂ − a₁ = 11 − 7 = 4
a₃ − a₂ = 15 − 11 = 4
a₄ − a₃ = 19 − 15 = 4
i.e., a_k+1 − a_k is same every time. Therefore, this is an AP with common difference as 4 and first term as 7.
\(S_n = \frac n2 [2a + (n-1)d]\)

\(S_{15} = \frac {15}{2 }[2(7) + (15-1)4]\)

\(S_{15} = \frac {15}{2} [14 + 56]\)

\(S_{15 }= \frac {15}{2} (70)\)
\(S_{15} = 15 × 35\)
\(S_{15} = 525\)

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(ii) a_n = 9 − 5n
a₁ = 9 − 5 × 1 = 9 − 5 = 4
a₂ = 9 − 5 × 2 = 9 − 10 = −1
a₃ = 9 − 5 × 3 = 9 − 15 = −6
a₄ = 9 − 5 × 4 = 9 − 20 = −11
It can be observed that
a₂ − a₁ = − 1 − 4 = −5
a₃ − a₂ = − 6 − (−1) = −5
a₄ − a₃ = − 11 − (−6) = −5
i.e., a_k+1 − a_k is same every time. Therefore, this is an A.P. with common difference as −5 and first term as 4.

\(S_n = \frac n2 [2a + (n-1)d]\)

\(S_{15 }= \frac {15}{2} [2(4) + (15-1)(-5)]\)

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

\(S_{15 }= \frac {15}{2} [8 - 70]\)

\(S_{15 }= \frac {15}{2} (-62)\)
\(S_{15} = 15 (-31)\)
\(S_{15 }= -465\)