Question

The sum of the 2nd and 7th term of an Ap is 30.If it's 15th term is 1 less than twice its8th term.Find the Ap.​

Answer 1

EXPLANATION.

Sum of n terms of an Ap.

An = a + ( n - 1 ) d

Sum of the 2nd and 7th term of an Ap

= 30.

$\rm \to \: {t_{2}} + t_{7} \: = 30 \: ......(1)$

=> a + d + a + 6d = 30

=> 2a + 7d = 30 ...... (1)

If 15th term is 1 less than twice it's 8th term.

$\rm \to \: t_{15} \: = 2( t_{8}) \: -1 \: \: .......(2)$

=> a + 14d = 2 ( a + 7d ) - 1

=> a + 14d = 2a + 14d - 1

=> a = 1

put the value of a = 1 in equation (1)

we get,

=> 2(1) + 7d = 30

=> 7d = 28

=> d = 4

Therefore,

First term = a = 1

Second term = a + d = 5

Third term = a + 2d = 9

Fourth term = a + 3d = 13

Sequence = 1,5,9,13.......

Answer 2

Answer:

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…..

Step-by-step explanation:

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