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CBSE 10th Standard Maths Subject Arithmetic Progressions Ncert Exemplar 1 Marks Questions With Solution 2021

Published on: 26/05/2021

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Questions:

Generic 2

10 x 2 = 20

If the common difference of an AP is 5, then what is a₁₈- a₁₃?

Which term of the AP 21, 42, 63, 84, ... is 210?

Find the 4th term from the end of an AP -11, -8, -5,....., 49.

In an AP, if d = -4, n =7 and \(a_n=4,\) then find the value of a first term.

If the n^th terms of the two A.P's. : 9, 7, 5, ...... and 24, 21, 18, .....are the same, find the value of n. Also, find that term.

Name the famous mathematician associated with finding the sum of the first 100 natural numbers.

If the common difference of an A.P. is 5, then find the value of a₁₈ - a₁₃ .

If 7 times the 7th term of A.P. is equal to 11 times the 11th term, then find the 18th term.

If the first term of an A.P. is -5 and the common difference is 2, then find the sum of the first 6 terms.

10.

Find the 6th term from the end of the A.P. -11, -8, -5, ...., 49

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Answers Key:

Generic 2

10 x 2 = 20

a₁₈- a₁₃ = (a + 17d) - (a + 12d) - (a + 12d)
= 5d = 25
[ \(\because\) d = 5 ]
= 25

Let nth term, \(a_n\) be 210.
Then, a + (n-1)d = 210 \([\because a_n=a+(n-1)d]\)
\(\Rightarrow\) 21 + (n-1) 21 = 210
\(\Rightarrow\) [\(\because\) a = 21, d =42 - 21 = 21, given]
\(\Rightarrow\) 21 + 21n - 21 = 210
\(\therefore\) n = 10

We know that, the nth term of an AP from the end is
\({a}_{n}=l-(n-1)d\) ...(i)
Here, l = last term = 49
and common difference, d = -8 - (-11)
= - 8 + 11 = 3
From Eq. (i), \({ a }_{ 4 }=49-(4-1)3=49-9=40\)

Given, d = - 4, n = 7 and \(a_n=4\)
We know that,
\({a}_{n}=a + (n-1)d\)
\(\Rightarrow\) 4 = a + (7 - 1) (-4)
\(\Rightarrow\) 4 = a + 6 (-4)
\(\Rightarrow\) 4+24 = a
\(\Rightarrow\) a = 28

Here, n term^th term of I A.P. = n^th term of II A.P.
\(\Rightarrow \) 9 + (n - 1) (- 2) = 24 + (n-1)(-3)
\(\Rightarrow \) 9 - 2n + 2 = 24 - 3n + 3
\(\Rightarrow \) n = 27 - 11 = 16
Hence, the value of n is 16.