The eighth term of an AP is half its second term and the eleventh term exceeds one-third of its fourth term by 1. Find the 15th term. - Mathematics

Sum

The eighth term of an AP is half its second term and the eleventh term exceeds one-third of its fourth term by 1. Find the 15th term.

Solution

Let a and d the first term and common difference of an AP, respectively

Now, by given condition, a_8 = 1/2 a_2

⇒ a + 7d = 1/2 (a + d)  ......[because a_n = a + (n - 1)s]

⇒ 2a + 14d = a + d

⇒ a + 13d = 0  ......(i)

And a_11 = 1/3 a_4 + 1 ......[Given]

⇒ a + 10d = 1/3 [a - 3d] + 1

⇒ 3a + 30d = a + 3d + 3

⇒ 2a + 27d = 3

From equations (i) and (ii)

2(– 13d) + 27d = 3

⇒ – 26d + 27d = 3

⇒ d = 3

From equation (i),

a + 13(3) = 0

⇒ a = – 39

∴ a15 = a + 14d

= – 39 + 14(3)

= – 39 + 42

= 3

Concept: Sum of First n Terms of an A.P.
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APPEARS IN

NCERT Mathematics Exemplar Class 10
Chapter 5 Arithematic Progressions
Exercise 5.4 | Q 3 | Page 57
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