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The eighth term of an AP is half its second term and the eleventh term exceeds one- third of its fourth term by 1. Then, the $15 t h$ term is 3.

State true (T) or false (F).

see full answer

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True

False

answer is A.

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Detailed Solution

Given,

a 8 = 12 a 2 and a 11 = 13 a 4 + 1.

For A.P., the first term = a, common difference = d and n = numbers of terms.
According to the question, the eighth term of an AP is half its second term i.e.,

a 8 = 12 a 2 a 8 = 12 a + 2 − 1 d a 8 = 12 a + d

And, eleventh term exceeds one-third of its fourth term by 1 thus,

a 11 = 13 a 4 + 1 a 11 = 13 a + 4 − 1 d + 1 a 11 = 13 a + 3 d + 1

To find no. of terms of an AP, let’s use,

a n = a + n − 1 d

For

a 8

, n = 8

a 8 = a + 8 − 1 d a 8 = a + 7 d .... 1

Substitute

12 a + d

for

a 8

from above expression (1), we get,

12 a + d = a + 7 d 2 a + 14 d = a + d a + 13 d = 0 .... 2

For

a 11

, n = 11

a 11 = a + 11 − 1 d a 11 = a + 10 d .... 3

Substitute

13 a + 3 d + 1

for

a 11

in the above (3),

13 a + 3 d + 1 = a + 10 d 3 a + 30 d = a + 3 d + 3 2 a + 27 d = 3 .... 4

.
Multiply equation (2) by 2,

2 a + 26 d = 0

.... 5

Now, by (4) and (5)
a = -39 and d = 3
For 15th term,

a n = a + n − 1 d

a 15 = − 39 + 15 − 1 3 a 15 = − 39 + 42 a 15 = 3

.
So, the 15th term is 3.
Therefore, the given statement, “The eighth term of an AP is half its second term and the eleventh term exceeds one- third of its fourth term by 1. Then, the

term is 3” is true.
Correct option is 1.

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