Question

sum of 4th and 8rh term of an AP is 24 and sum of 6th and 10 th term is 34
find first term and common diffrence​

Answer 1

Given:-

• Sum of 4th and 8th term of an AP is 24 and Sum of 6th and 10th term is 34.

To Find:-

• First Term

• Common Difference

Formula Used:-

• ${\boxed{\bf{a_n=a+(n-1)d}}}$

Here,

• $\bf a_n$ = General term

• a = First Term

• n = No. of Terms

• d = Common Difference

Solution:-

According to question,

$\bf :\implies\:a_4+a_8=24$

Solving it further,

$\sf :\implies\:a +(4-1)d+a+(8-1)d=24$

$\sf :\implies\:a +3d+a+7d=24$

$\sf :\implies\:2a +10d=24$

$\sf :\implies\:a +5d=12$

$\sf :\implies\:a =12-5d$ ___{1}

Also,

$\bf :\implies\:a_6+a_{10}=34$

Solving it further,

$\sf :\implies\:a +(6-1)d+a+(10-1)d=34$

$\sf :\implies\:a +5d+a+9d=34$

$\sf :\implies\:2a +14d=34$

$\sf :\implies\:a +7d=17$

Putting value of a in it,

$\sf :\implies\:12-5d +7d=17$

$\sf :\implies\:12+2d=17$

$\sf :\implies\:2d=17-12$

$\sf :\implies\:d=\dfrac{5}{2}$

$\bf :\implies\:d=2.5$

Putting value of d in {1},

$\sf :\implies\:a =12-5\times2.5$

$\sf :\implies\:a =12-12.5$

$\bf :\implies\:a =-0.5$

Hence, The First Term of an A.P is -0.5 and the Common Difference 2.5.

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Answer 2

Given, in an A.P :

• The sum of 4ᵗʰ and 8ᵗʰ terms of an A.P. is 24

⇒ a₄ + a₈ = 24

And, we know that :

• aₙ = a + (n – 1)d

➝ [a + (4-1)d] + [a + (8-1)d] = 24

➝ 2a + 10d = 24

➝ a + 5d = 12….①

Also given that,

• the sum of the 6ᵗʰ and 10ᵗʰ terms is 34

⇒ a₆ + a₁₀ = 34

⇒ [a + 5d] + [a + 9d] = 34

⇒ 2a + 14d = 34

⇒ a + 7d = 17….②

Subtracting ① form ②, We have

➻ a + 7d – (a + 5d) = 17 – 12

➻ 2d = 5 d = 5/2

Using d in ① we get,

➻ a + 5(5/2) = 12

➻ a = 12 – 25/2

➻ a = -1/2

⚘ Therefore,

• Hence, the first term is -1/2 and the common difference is 5/2.

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⍟ Know More :

• ➭ aₙ = General Term

• ➭ a = First Term

• ➭ n = No. of Terms

• ➭ d = Common Difference

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