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) an = 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., ak+1 − ak 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) an = 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., ak+1 − ak is same every time. Therefore, this is an A.P. with common difference as −5 and first term as 4.

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

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

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

$S_{15 }= \frac {15}{2} [8 - 70]$

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