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Sum of N Terms of an AP

Question 23 I...

Question

Question 23
If $a_{n} = 3 - 4 n$ , then show that $a_{1}, a_{2}, a_{3}, \dots$ form an AP. Also, find $S_{20}$ .

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Solution

Given: nth term of the series, $a_{n} = 3 - 4 n$

Put $n = 1$ , $a_{1} = 3 - 4 (1) = 3 - 4 = - 1$
Put $n = 2$ , $a_{2} = 3 - 4 (2) = 3 - 8 = - 5$
Put $n = 3$ , $a_{3} = 3 - 4 (3) = 3 - 12 = - 9$
Put $n = 4$ , $a_{4} = 3 - 4 (4) = 3 - 16 = - 13$
We see that,
$a_{2} - a_{1} = - 5 - (- 1) = - 5 + 1 = - 4$
$a_{3} - a_{2} = - 9 - (- 5) = - 9 + 5 = - 4$
$a_{4} - a_{3} = - 13 - (- 9) = - 13 + 9 = - 4$
$a_{2} - a_{1} = a_{3} - a_{2} = a_{4} - a_{3} = \dots = - 4$
Since, the each successive term of the series has the same difference, it forms an AP.
We know that, sum of n terms of an AP,
$S_{n} = \frac{n}{2} [2 a + (n - 1) d]$
$∴$ Sum of 20 terms of the AP,
$S_{20} = \frac{20}{2} [2 (- 1) + (20 - 1) (- 4)]$
$= 10 (- 2 + (19) (- 4)) = 10 (- 2 - 76)$
$= 10 \times - 78 = - 780$

Hence, $S_{20} = - 780$

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Similar questions

a_{1}, a_{2}, a_{3}, . . . . . . . a_{n}

form an AP where

a_{n} = 3 + 4 n

a_{1}, a_{2}, . . ., a_{n}

form an AP where

a_{n}

is defined as below :
(i)

a_{n} = 3 + 4 n

a_{n} = 9 - 5 n

Also find the sum of the first

15

terms in each case.

Q. Question 2
Justify whether it is true to say that

- 1, \frac{- 3}{2}, - 2, \frac{5}{2}, \dots

a_{2} - a_{1} = a_{3} - a_{2}

.

\frac{a_{1}}{a_{1}} + \frac{a_{2}}{a_{2}} + \frac{a_{3}}{a_{3}} + \dots \frac{a_{n}}{a_{n}} = \frac{1}{1} + \frac{1}{a_{1}} + \frac{a_{1}}{a_{2}} + \frac{a_{2}}{a_{3}} + \dots \frac{a_{n - 2}}{a_{n - 1}}

a_{2} = 13;

a_{4} = 3

a_{1}, a_{3}

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Sum of N Terms of an AP

Standard X Mathematics

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