The 26th, 11th and the last term of an AP are 0, 3 and -⅕ respectively. Find the common difference and the number of terms

Solution:

Consider the first term, common difference and number of terms of an AP are a, d and n, respectively.

If last term of an AP is known,

l = 8 + (11-1 )d …………. (i)

So nth term of an AP is

T_n = a + (n - 1)d …………. (ii)

We know that,

26th term of an AP = 0

T₂₆ = a + (26 - 1 )d = 0 [from Equation (i)]

8 + 25d = 0 …………. (iii)

11th term of an AP = 3

T₁₁ = s + (11 - 1)d = 3 [from Equation (ii)]

8 + 10d = 3 ……………… (iv)

Last term of an AP = -⅕

l = a + (n - 1 )d [from Equation (i)]

-1/5 = a + (n - 1 )d ………… (v)

Now, subtracting Equation (iv) from Equation (iii),

15 d = - 3

d = -⅕

Substitute the value of d in Equation (iii),

a + 25(-⅕) = 0

a - 5 = 0

8 = 5

Substitute the value of a, d in Equation (v), we get

-1/5 = 5 + (n - 1)(-1/5)

-1 = 25 - (n - 1)

-1 = 25 - n + 1

n = 25 + 2 = 27

Therefore, the common difference and number of terms are -1/ 5 and 27.

✦ Try This: The 25th, 10th and the last term of an AP are 3, 5 and -½ respectively. Find the common

difference and the number of terms

☛ Also Check: NCERT Solutions for Class 10 Maths Chapter 5

NCERT Exemplar Class 10 Maths Exercise 5.3 Problem 6

Summary:

The 26th, 11th and the last term of an AP are 0, 3 and -⅕ respectively. The common difference and number of terms are -1/ 5 and 27.

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