# The 26th, 11th and the last term of an AP are 0, 3 and -15, respectively. Find the common difference and the number of terms. - Mathematics

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

#### Solution

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

We know that, if last term of an AP is known, then

l = 8 + (11 – 1)d   ........(i)

And nth term of an AP is

Tn = a + (n – 1)d   .........(ii)

Given 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)

And last term of an AP = -1/5

⇒ l = a + (n – 1 )d   .......[From equation (i)]

⇒ -1/5 = a + (n – 1 )d   ........(v)

Now, subtracting equation (iv) from equation (iii),
15d = – 3

⇒ d = - 1/5

Put the value of d in equation (iii), we get

a + 25(-1/5) = 0

⇒ a – 5 = 0

⇒ 8 = 5

Now put 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

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

Concept: Arithmetic Progression
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#### APPEARS IN

NCERT Mathematics Exemplar Class 10
Chapter 5 Arithematic Progressions
Exercise 5.3 | Q 6 | Page 52
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