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CBSE 10th Standard Maths Subject Arithmetic Progressions Ncert Exemplar 3 Marks Questions 2021

Published on: 26/05/2021

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Questions:

Generic 3

5 x 3 = 15

Find the sum of those integers between 1 and 500, which are multiples of 2 as well as of 5.

Find the sum of first 17 terms of an AP, where 4th and 9th terms are -15 and - 30, respectively.

If sum of first 6 terms of an AP is 36 and that of the first 16 terms is 256, then find the sum of first 10 terms.

The sum of the first five terms and the sum of the first seven terms of an AP is 167. If the sum of the first ten terms of this AP is 235. then find the sum of its first twenty terms.

Kanika was given her pocket money on Jan 1st, 2008. She puts Rs 1 on day, 1, Rs 2 on day 2, Rs 3 on day 3 and continued doing so till the end of the month, from this money into her piggy back she also spent Rs 204 of her pocket money and found that at the end of the month she still had Rs 100 with her. How much was her pocket money for the month?

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Answers Key:

Generic 3

5 x 3 = 15

Consider 10, 20, 30, 40, ... 490. According to question
490 = 10 + (n - 1) x 10
[\(\because\) a = 10, an = 490]
and \(S_{n}=\frac{n}{2}[2 \times 10+(n-1) \times 10]\)
On solving, we get n = 49.
= 12,250

- 510

100

Given, S₅ + S₇ = 167 and S₁₀ = 235
\(\Rightarrow 12 a+31 d=167 \text { and } 2 a+9 d=47\)
Now, find a and d, and then S₂₀
S₂₀ = 970

Let her pocket money be Rs x. Now, she takes Rs 1 on day 1, Rs 2 on day 2, Rs 3 on day 3 and so on till the end of the month, from this money.
i.e. 1 + 2 + 3 + 4 + ..... + 31
which form an AP, in which number of terms is 31 and first term
\(\therefore\) Sum of first terms = S ₃₁
Sum of n terms, \({ S }_{ n }=\frac { n }{ 2 } \left[ 2a+\left( n-1 \right) d \right] \)
\(\therefore\) \({ S }_{ 31 }=\frac { 31 }{ 2 } \left[ 2\times 1+\left( 31-1 \right) \times 1 \right] \)
\(=\frac { 31 }{ 2 } \left( 2+30 \right) \)
\(=\frac { 31\times 32 }{ 2 } \)
\(=31\times 16=496\)
So, Kanika takes Rs 496 till the end of the month from this money.
Also, she spent Rs 204 of her pocket money and found that at the end of the month, she still has Rs 100 with her.
Now, according to the condition
\(\left( x-496 \right) -204=100\)
\(\Rightarrow\) x - 700 = 100
\(\Rightarrow\) x = Rs 800
Hence, Rs 800 was her pocket money for the month.