Question

The sum of the 2nd and the 7th terms of an AP is 30. If its 15th term is
1 less than twice its 8th term, find the AP.
(CBSE 2014

Answer 1

Answer:

AP is 1, 5, 9, 13 if term is 1 less than twice of its 8 term.

Step-by-step explanation:

Solution:

Addition of and term of AP = 30

(A+ D) + (A + 6D) = 30

Sum = 2A + 7D = 30.

Now as the question says the subtraction of 15th term and 2(8th term) = 1, that will be  

(A + 14D) – 2(A + 7D) = 1

(A + 14D) – 2(A + 7D) = 1

A – 2A = 1

Value of A = 1

First term = 1,  

Difference can be found by substituting value of A in (A+ D) + (A + 6D) = 30 ) we get,

(1+ D) + (1 + 6D) = 30

2 + 7D = 30

D = 4.  

With both the values of A and D, the Arithmetic Progression as 1, 5, 9, 13…..

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Answer 2

Answer:

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Step-by-step explanation:

AP is 1, 5, 9, 13 if \bold{15^{th}} term is 1 less than twice of its \bold{8^{th}} term.

Solution:

Addition of 2^{nd} and 7^{th} term of AP = 30

(A+ D) + (A + 6D) = 30

Sum = 2A + 7D = 30.

Now as the question says the subtraction of 15th term and 2(8th term) = 1, that will be

(A + 14D) – 2(A + 7D) = 1

(A + 14D) – 2(A + 7D) = 1

A – 2A = 1

Value of A = 1

First term = 1,

Difference can be found by substituting value of A in (A+ D) + (A + 6D) = 30 ) we get,

(1+ D) + (1 + 6D) = 30

2 + 7D = 30

D = 4.

With both the values of A and D, the Arithmetic Progression as 1, 5, 9, 13…..

From Divya

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