Question

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$If \: S_{3} = -4, \: and \\ S_{4} = -3 ,\:in \:A.P \:then\ \prove\:that \:S_{7} = 7 \:A.P$​

Answer 1

$\huge{\underline{\underline{\mathfrak{Answer}}}}$

➡Refer to the attachment ☺

* Actually the procedure is quite lengthy and I do not have a shorter method.

$\huge{\underline{\underline{\blue{\mathfrak{Formula Used }}}}}$

(1) Sn=n/2 (2a+(n-1) d)

(2) Elimination method for the value of a and (d) .

$\huge{\underline{\underline{\red{\mathfrak{Thank You }}}}}$

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# BAL

Answer 2

Hey there!

General Formula of Sn:

$S_n = \dfrac{n}{2}\left[ 2a + (n-1)d]$

Given,

$S_3 = -4\\\\\dfrac{n}{2}[2a + (n-1)d] = -4\\\\\dfrac{3}{2}(2a + 2d) = -4\\\\3(2a + 2d) =-8\\\\\boxed{6a+ 6d = -8}..........(i)$

And,

$S_4 = -3\\\\\dfrac{n}{2}[2a + (n-1)d] = -3\\\\\dfrac{4}{2}(2a + 3d) = -3\\\\\boxed{4a + 6d = -3}..........(ii)$

Subtracting Equation (ii) from (i):

$6a + 6d = -8\\4a + 6d = -3\\-~~~- ~~~~+\\=========\\2a = -5\\\boxed{a=\frac{-5}{2}}$

Putting value of a in eq. (ii);

$4a + 6d = -3\\\\4 \times \dfrac{-5}{2} + 6d = -3\\\\-10 + 6d = - 3\\\\6d = -3 + 10\\\\\boxed{d = \dfrac{7}{6}}$

We have to prove that S₇ = 7:

Taking L.H.S:

$S_7\\\\=\dfrac{n}{2} [2a + (n-1)d]\\\\=\dfrac{7}{2} \left[2 \times \left(\dfrac{-5}{2}\right) +(7-1)\times \dfrac{7}{6}\right]\\\\=\dfrac{7}{2}(-5 + 7)\\\\=7\\\\=R.H.S$

Hence Proved!