Question

the sum of the 4th term and 8th term of an AP is 24 and the sum of the 6th and 10th is 44. Find first three terms of the AP​

Answer 1

$\huge\red{\underline{\underline{\pink{Ans}\red{wer:-}}}}$

$\sf{First \ three \ terms \ are \ -13, \ -8 \ and}$

$\sf{-3 \ respectively.}$

$\huge\orange{Given:}$

$\sf{In \ an \ A.P.,}$

$\sf{\implies{t4+t8=24}}$

$\sf{\implies{t6+t10=44}}$

$\huge\pink{To \ find:}$

$\sf{First \ three \ terms \ of \ the \ A.P.}$

$\huge\green{\underline{\underline{Solution:}}}$

$\sf{tn=a+(n-1)d...formula}$

$\sf{\implies{t4+t8=24}}$

$\sf{a+3d+a+7d=24}$

$\sf{\therefore{2a+10d=24}}$

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

$\sf{\therefore{a+5d=\frac{24}{2}}}$

$\sf{\implies{a+5d=12...(1)}}$

$\sf{\implies{t6+t10=44}}$

$\sf{a+5d+a+9d=44}$

$\sf{\therefore{2a+14d=44}}$

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

$\sf{\therefore{a+7d=\frac{44}{2}}}$

$\sf{\implies{a+7d=22...(2)}}$

$\sf{Subract \ equation \ (1) \ from \ equation (2)}$

$\sf{a+7d=22}$

$\sf{-}$

$\sf{a+5d=12}$

________________

$\sf{2d=10}$

$\sf{\therefore{d=\frac{10}{2}}}$

$\sf{\implies{d=5}}$

$\sf{Substitute \ d=5 \ in \ equation \ (1)}$

$\sf{a+5(5)=12}$

$\sf{\therefore{a+25=12}}$

$\sf{\therefore{a=12-25}}$

$\sf{\implies{a=-13}}$

$\sf{First \ three \ terms \ of \ A.P. \ are:}$

$\sf{a, \ a+d \ and \ a+2d}$

$\sf{\therefore{t1=-13, \ t2=-13+5=-8 \ and}}$

$\sf{t3=-13+2(5)=-13+10=-3}$

$\sf\purple{\tt{\therefore{First \ three \ terms \ are \ -13, \ -8 \ and}}}$

$\sf\purple{\tt{-3 \ respectively.}}$

Answer 2

$\large{\underline{\bf{\pink{Given:-}}}}$

• ✭ Sum of 4th term and 8th term

⠀⠀⠀⠀⠀⠀⠀ = 24

• ✭ sum of 6th and 10th term = 44

$\large{\underline{\bf{\pink{To\:Find:-}}}}$

• ✭ we have to find 1st three terms.

$\huge{\underline{\bf{\purple{Solution:-}}}}$

we know that :-

$\longmapsto\rm\:a_ 4 = a + 3d\\ \\ \: \longmapsto\rm \: a_ 8 = a + 7d \\ \\ \longmapsto\rm \: a_ 6 = a + 5d \\ \\ \longmapsto\rm \: a_ {10} = a + 9d$

$\longmapsto \rm\:T_4+T_8=24\: \\ \\ \longmapsto \rm\:(a + 3d) +(a + 7d) = 24 \\ \\\longmapsto \rm\:2a + 10d = 24 \\ \\ \rm\: divide \: both \: side \: by \: 2 \\ \\\longmapsto \rm\:a + 5d = 12 \: \: \: \: \: ..........(i)$

$\longmapsto \rm\:and\:\:T_6+T_{10}=44\: \\ \\\longmapsto \rm\:(a + 5d) + (a + 9d) = 44 \\ \\\longmapsto \rm\:2a + 14d = 44 \\ \\ \rm\:divide \: both \: side \: by \: 2 \\ \\ \longmapsto \rm\:a + 7d = 22 \: \: \: \:......... (ii)$

$\rm\:a + 5d = 12 \\ \rm \: a + 7 d = 22\\ - \: \: \: - \: \: \: \: \: \: \: - \\ \underline{ \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: } \\ \\ \rm \: \: \: \: \: - 2d \: \: = - 10 \\ \rm \: \: \: \: \: \: \: d = \cancel \frac{ - 10}{ - 2} \\ \\ \rm \: \: \: \: \: \: \: d = 5$

putting value of d in equation (i).

$\longmapsto \rm\:a + 5d = 12\:\\\\\longmapsto \rm\:a+5\times5=12\\\\\longmapsto \rm\:a+25=12\\\\\longmapsto \rm\:a=-13$

Now,

first three terms are :-

$\longmapsto \rm\:a = - 13\: \\ \\ \longmapsto \rm\:a + d = - 13 + 5 \\ \\ = > \rm\: -8 \\ \\\longmapsto \rm\: a + 2d = - 13 + 2 \times 5 \\ \\\longmapsto \rm\: - 13 + 10 \\ \\ \longmapsto \rm\: - 3$

Hence first three terms of AP are :-

-13 , -8 , -3

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