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Sum of n Terms in AP

Question 23If...

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 – 4n
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

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

form an AP where

a_{n} = 3 + 4 n

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

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

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form an AP where

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Also find the sum of the first

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\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}}

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Justify whether it is true to say that

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Sum of n Terms in AP

Standard XI Mathematics

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