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Home » 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.
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.
Arithmetic Progression | CBSE | Class 10 | Exercise 5.4 | Maths | NCERT Exemplar
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:

We all know that,

AP’s first term = a

AP’s common difference = d

AP’s nth term, {{a}_{n}}~=\text{ }a\text{ }+\text{ }\left( n\text{ }\text{ }1 \right)d

According to the question,

{{a}_{s}}~=\text{ }{\scriptscriptstyle 1\!/\!{ }_2}\text{ }{{a}_{2}}

2{{a}_{8}}~=\text{ }{{a}_{2}}

2\left( a\text{ }+\text{ }7d \right)\text{ }=\text{ }a\text{ }+\text{ }d

2a\text{ }+\text{ }14d\text{ }=\text{ }a\text{ }+\text{ }d

a\text{ }=13d…(1)

Also,

{{a}_{11}}~=\text{ }1/3\text{ }{{a}_{4}}~+\text{ }1

3\left( a\text{ }+\text{ }10d \right)\text{ }=\text{ }a\text{ }+\text{ }3d\text{ }+\text{ }3

3a\text{ }+\text{ }30d\text{ }=\text{ }a\text{ }+\text{ }3d\text{ }+\text{ }3

2a\text{ }+\text{ }27d\text{ }=\text{ }3

On substituting a = -13d in the equation,

2\text{ }\left( -\text{ }13d \right)\text{ }+\text{ }27d\text{ }=\text{ }3

d=3

Then,

a\text{ }=\text{ }13\left( 3 \right)=\text{ }39

Now,

{{a}_{15}}~\text{ }=\text{ }a\text{ }+\text{ }14d

=39\text{ }+\text{ }14\left( 3 \right)

=39\text{ }+\text{ }42

=3

As a result 15th term is 3.

i

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