[tex]\red {13}[/tex]
[tex] If \: S_{3} = -4, \: and \\ S_{4} = -3 ,\:in \:A.P \:then\ \prove\:that \:S_{7} = 7 \:A.P[/tex]
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[tex]\red {13}[/tex]
[tex] If \: S_{3} = -4, \: and \\ S_{4} = -3 ,\:in \:A.P \:then\ \prove\:that \:S_{7} = 7 \:A.P[/tex]
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[tex] \huge{\underline{\underline{\mathfrak{Answer}}}}[/tex]
➡Refer to the attachment ☺
* Actually the procedure is quite lengthy and I do not have a shorter method.
[tex] \huge{\underline{\underline{\blue{\mathfrak{Formula Used }}}}}[/tex]
(1) Sn=n/2 (2a+(n-1) d)
(2) Elimination method for the value of a and (d) .
[tex] \huge{\underline{\underline{\red{\mathfrak{Thank You }}}}}[/tex]
❤Ⓗⓞⓟⓔ ⓘⓣ ⓗⓔⓛⓟⓢ ❤
# BAL
Verified answer
Hey there!
General Formula of Sn:
[tex]S_n = \dfrac{n}{2}\left[ 2a + (n-1)d][/tex]
Given,
[tex]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)[/tex]
And,
[tex]S_4 = -3\\\\\dfrac{n}{2}[2a + (n-1)d] = -3\\\\\dfrac{4}{2}(2a + 3d) = -3\\\\\boxed{4a + 6d = -3}..........(ii)[/tex]
Subtracting Equation (ii) from (i):
[tex]6a + 6d = -8\\4a + 6d = -3\\-~~~- ~~~~+\\=========\\2a = -5\\\boxed{a=\frac{-5}{2}}[/tex]
Putting value of a in eq. (ii);
[tex]4a + 6d = -3\\\\4 \times \dfrac{-5}{2} + 6d = -3\\\\-10 + 6d = - 3\\\\6d = -3 + 10\\\\\boxed{d = \dfrac{7}{6}}[/tex]
We have to prove that S₇ = 7:
Taking L.H.S:
[tex]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[/tex]
Hence Proved!