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# Question 11. The 26th, 11th and the last term of an A.P. are 0, 3 and -1/5, respectively. Find the common difference and the number of terms.

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One of the most important and exam oriented question from Chapter name- Arithmetic Progression
Class 10th
Chapter number- 9
Exercise :- 9.4
This type of question has been asked in previous years exams.

In this question we have been given that the 26th, 11th and the last term of an A.P. are 0, 3 and -1/5, respectively.

Now we have to find the common difference and the number of terms.

CBSE DHANPAT RAI publication
Understanding CBSE Mathematics
Question 11

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1. Given that,

a26 = 0, a11 = 3 and an (last term) = -1/5 of an A.P.

As we know that, to find nth term in an A.P = a + (n – 1)d

Therefore,

a26 = a + (26 – 1)d

a + 25d = 0 ——–(1)

a11 = a + (11 – 1)d

a + 10d = 3 ———(2)

Solving (1) and (2),

(1) – (2)

a + 25d – (a + 10d) = 0 – 3

15d = -3

d = -1/5

Using d in (1), we get

a + 25(-1/5) = 0

a = 5

Now, given that the last term is -1/5

5 + (n – 1)(-1/5) = -1/5

5 + -n/5 + 1/5 = -1/5

25 – n + 1 = -1

n = 27

Hence, the A.P has 27 terms and its common difference is -1/5.

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