Question

the sum of first five term of an AP is equal to 15 more than twice the next five terms and 4 th term is 15 then find 4th term​

Answer 1

Given:

The sum of the first five-term of an AP is equal to 15 more than twice the next five terms and the first term is 15

To find:

The 4th term​

Solution:

The first term of the given A.P., a = 15

We know,

$\boxed{\bold{S_n= \frac{n}{2} [2a+ (n-1)d] }}$

The sum of 1st 5-term of the A.P. is,

S₅ = $\frac{5}{2} [(2\times 15)+ (5-1)d] = \frac{5}{2} [30+ 4d] = 75 + 10d$

The sum of 1st 10-term of the A.P. is,

S₁₀ = $\frac{10}{2} [(2\times 15)+ (10-1)d] =5 [30+ 9d] = 150 + 45d$

∴ The sum of the next 5-terms of an A.P. is,

= S₁₀ - S₅

= 150 + 45d - (75 + 10d)

= 150 + 45d - 75 - 10d

= 75 + 35d

The sum of the first five-term of an AP is equal to 15 more than twice the next five terms, so we can form an equation as,

[Sum of 1st 5-terms] = 15 + 2[Sum of next 5-terms]

$\implies 75 + 10d = 15 + 2(75 + 35d)$

$\implies 75 + 10d = 15 + 150 + 70d$

$\implies 10d - 70d = 15 + 150 - 75$

$\implies - 60d = 90$

$\implies - d = \frac{9}{6}$

$\implies \bold{d = - \frac{3}{2}}$

∴ The 4th term of the A.P. "t₄" is,

= a + (n - 1)d

= 15 + (4 - 1)($-\frac{3}{2}$)

= 15 + ( 3 × $-\frac{3}{2}$)

= 15 + 4.5

= $\bold{19.5}$

Thus, the 4th term of the given A.P. is → 19.5.

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