Question

Q(4) The sum of 4th and 8th terms of an A.P. is 24 and the sum of the 6th

and 10th terms is 44. Find the first two terms of the A.P.​

Answer 1

$\displaystyle\large\underline{\sf\red{Given}}$

✭ Sum of 4th and 8th term = 24

✭ Sum of 6th and 10 term = 44

$\displaystyle\large\underline{\sf\blue{To \ Find}}$

◈ First two terms of A.P?

$\displaystyle\large\underline{\sf\gray{Solution}}$

We shall here primary use the formula,

$\displaystyle\underline{\boxed{\sf a_n = a+(n-1)d}}$

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$\underline{\bigstar\:\textsf{According to the given Question :}}$

➢ $\displaystyle\sf a_4 = a_1 + 3d$

➢ $\displaystyle\sf a_8 = a_1 + 7d$

By given,

➳ $\displaystyle\sf a_4 + a_8 = 24$

➳ $\displaystyle\sf a_1 + 3d + a_1 + 7d = 24$

➳ $\displaystyle\sf 2a_1 + 10d = 24$

➳ $\displaystyle\sf a_1 + 5d = 12\:\:\: -eq(1)$

Also,

»» $\displaystyle\sf a_6 = a_1 + 5d$

»» $\displaystyle\sf a_{10} = a_1 + 9d$

By given

›› $\displaystyle\sf a_6 + a_{10} = 44$

›› $\displaystyle\sf a_1 + 5d + a_1 + 9d = 44$

›› $\displaystyle\sf 2a_1 + 14d = 44$

›› $\displaystyle\sf a_1 + 7d = 22\:\:\:-eq(2)$

Solve and 2 by elimination method

➠ $\displaystyle\sf -2d = -10$

➠ $\displaystyle\sf \orange{d = 5}$

Substitute in eq(2)

➝ $\displaystyle\sf a_1 + 35 = 22$

➝ $\displaystyle\sf \purple{a_1 = -13}$

➝ $\displaystyle\sf a_2 = a_1 + d$

➝ $\displaystyle\sf \pink{a_2 = -8}$

$\displaystyle\therefore\:\underline{\sf First \ term = -13, Second \ term = -8}$

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Answer 2

Given:

$\bull$ Sum of 4th and 8th term of an A.P. = 24

$\bull$ Sum of 6th and 10th term of an A.P. = 44

Find:

$\bull$ First two terms of A.P.

Solution:

we, know that

$\underbrace{\underline{\boxed{\pink{\sf a_n = a + (n - 1)d}}}}$

So,

$\underline{\underline{\boxed{\textrm{ \purple{According \:to \: Question}}}}}$

For, 4th term

$\sf \to a_n = a + (n - 1)d$

$\sf \to a_4 = a + (4 - 1)d$

$\sf \to a_4 = a + (3)d$

$\sf \to a_4 = a + 3d$

Now, For 8th term

$\sf \to a_n = a + (n - 1)d$

$\sf \to a_8 = a + (8 - 1)d$

$\sf \to a_8 = a + (7)d$

$\sf \to a_8 = a + 7d$

Now, Sum of 4th and 8th term is 24.

$\sf \therefore a_4 + a_8 = 24$

$\sf \to (a + 3d) + (a + 7d) = 24$

$\sf \to a + 3d + a + 7d = 24$

$\sf \to 2a + 10d = 24$

$\sf \to 2(a + 5d) = 24$

$\sf \to a + 5d = \cancel{\dfrac{24}{2}} = 12$

$\sf \to a + 5d= 12$

$\sf \to a= 12 - 5d....... \{1\}$

_____________________

Now, for 6th term

$\sf \to a_n = a + (n - 1)d$

$\sf \to a_6 = a + (6 - 1)d$

$\sf \to a_6 = a + (5)d$

$\sf \to a_6 = a + 5d$

For 10th term

$\sf \to a_n = a + (n - 1)d$

$\sf \to a_{10} = a + (10 - 1)d$

$\sf \to a_{10} = a + (9)d$

$\sf \to a_{10} = a + 9d$

So, sum of 6th and 10th term is 44.

$\sf \therefore a_6 + a_{10} = 44$

$\sf \to (a + 5d) + (a + 9d) = 44$

$\sf \to a + 5d + a + 9d = 44$

$\sf \to 2a + 14d = 44$

$\sf \to 2(a + 7d) = 44$

$\sf \to a + 7d= \cancel{\dfrac{44}{2}} = 22$

$\sf \to a + 7d= 22....... \{2\}$

_____________________

Now, use the eq{1} in eq{2} we, get

$\sf \longrightarrow a + 7d = 22$

$\sf \longrightarrow (12 - 5d)+ 7d = 22$

$\sf \longrightarrow 12 - 5d+ 7d = 22$

$\sf \longrightarrow 12 + 2d = 22$

$\sf \longrightarrow 2d = 22 - 12$

$\sf \longrightarrow 2d = 10$

$\sf \longrightarrow d = \cancel{\dfrac{10}{2} } = 5$

$\sf \longrightarrow d = 5$

So, d = 5

Now, use this value of d in eq(1)

$\sf \to a= 12 - 5d$

$\sf \to a= 12 - 5(5)$

$\sf \to a= 12 - 25$

$\sf \to a= - 13$

So, a = -13

_____________________

$\sf\therefore f1st \: term \: of \: series = - 13$

$\sf 2nd \: term \: of \: series = a + d \\ \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \sf = - 13 + 5 \\ \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \sf = - 8$

Hence, F1st two terms of A.P. are:- -13,-8