Question

do it fastr........ .,...​

Answer 1

$\underline{\sf Given :-}\\ \\\bullet\sf S_1=S_n \ \ ; \ \bullet\ S_2=S_{2n}\ \ ; \ \bullet S_3=S_{3n}$

$\underline{\sf To Prove :-}\\ \\\bullet\sf S_3=3(S_2-S_1)$

$\underline{\bigstar{\sf Solution:-}}$

We know the formula for the sum of nth term of an AP

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

Now ,

$\dashrightarrow\sf S_3=S_{3n} = \dfrac{3n}{2}[2a+(3n-1)d]\\ \\ \\ \dashrightarrow\sf S_2=S_{2n} = \dfrac{2n}{2}[2a+(2n-1)d]\\ \\ \\ \dashrightarrow\sf S_1=S_n = \dfrac{n}{2}[2a+(n-1)d]$

$\implies{\red{\sf S_3=\dfrac{3n}{2}[2a+(3n-1)d]-----eq.(i)}}$

★ Now First find the value of -

$\boxed{\rm\ S_2-S_1}$

$\dashrightarrow\sf \bigg[ \dfrac{2n}{2}[2a+(2n-1)d] \bigg] - \bigg[ \dfrac{n}{2}[2a+(n-1)d]\bigg]\\ \\ \\\dashrightarrow\sf \bigg[ \dfrac{2n[2a+(2n-1)d]}{2}\bigg] - \bigg[ \dfrac{n[2a+n-1)d]}{2}\bigg]\\ \\ \\\dashrightarrow\sf \bigg[ \dfrac{4na+4n^2 d-2nd}{2}\bigg] -\bigg[\dfrac{2na+n^2d-nd}{2}\bigg] \\ \\ \\ \dashrightarrow\sf \bigg[\dfrac{4na+4n^2d-2nd -2na-n^2d+nd}{2}\bigg] \ \ \ \ \big(\therefore \ By \ taking \ L.C.M \big)\\ \\ \\ \dashrightarrow\sf \bigg[\dfrac{(4na-2na)+(4n^2d-n^2d)+(-2nd+nd)}{2}\bigg]\\ \\ \\ \dashrightarrow\sf \bigg[\dfrac{2na+3n^2d-nd}{2}\bigg]\\ \\ \\ \dashrightarrow\sf \bigg[\dfrac{ n(2a+3nd-d)}{2}\bigg]\ \ \ \ \ \ \big(taking\ 'n' \ as \ common \big)\\ \\ \\ \dashrightarrow\sf \dfrac{n[2a+(3n-1)d]}{2}\\ \\ \\ \dashrightarrow\sf \dfrac{n}{2}[2a+(3n-1)d]$

Now we have ,

$\dashrightarrow\sf S_2-S_1= \dfrac{n}{2}[2a+(3n-1)d]\\ \\ \\ \bullet\sf Multiplying \ both \ side \ by \ (3) \\ \\ \\ \dashrightarrow\sf 3\big(S_2-S_1\big)= 3\bigg(\dfrac{n}{2}[2a+(3n-1)d]\bigg)\\ \\ \\ \dashrightarrow{\red{\sf 3\big(S_2-S_1\big)= \dfrac{3n}{2}[2a+(3n-1)d]----- eq.\ (ii)}}$

★ From eq. (i) and eq.(ii) we can write -

$\underline{\boxed{\sf S_3=3(S_2-S_1)}}$

$\underline{\underline{\dag{\sf \ \ \ \ Hence \ Proved\ \ \ \ \ \ }}}$

Answer 2

★ $\underline{\sf Given :-}\\ \\\bullet\sf S_1=S_n \ \ ; \ \bullet\ S_2=S_{2n}\ \ ; \ \bullet S_3=S_{3n}$

$\underline{\sf To Prove :-}\\ \\\bullet\sf S_3=3(S_2-S_1)$

$\underline{\bigstar{\sf Solution:-}}$

We know the formula for the sum of nth term of an AP

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

Now ,

$\dashrightarrow\sf S_3=S_{3n} = \dfrac{3n}{2}[2a+(3n-1)d]\\ \\ \\ \dashrightarrow\sf S_2=S_{2n} = \dfrac{2n}{2}[2a+(2n-1)d]\\ \\ \\ \dashrightarrow\sf S_1=S_n = \dfrac{n}{2}[2a+(n-1)d]$

$\implies{\red{\sf S_3=\dfrac{3n}{2}[2a+(3n-1)d]-----eq.(i)}}$

★ Now First find the value of -

$\boxed{\rm\ S_2-S_1}$

$\dashrightarrow\sf \bigg[ \dfrac{2n}{2}[2a+(2n-1)d] \bigg] - \bigg[ \dfrac{n}{2}[2a+(n-1)d]\bigg]\\ \\ \\\dashrightarrow\sf \bigg[ \dfrac{2n[2a+(2n-1)d]}{2}\bigg] - \bigg[ \dfrac{n[2a+n-1)d]}{2}\bigg]\\ \\ \\\dashrightarrow\sf \bigg[ \dfrac{4na+4n^2 d-2nd}{2}\bigg] -\bigg[\dfrac{2na+n^2d-nd}{2}\bigg] \\ \\ \\ \dashrightarrow\sf \bigg[\dfrac{4na+4n^2d-2nd -2na-n^2d+nd}{2}\bigg] \ \ \ \ \big(\therefore \ By \ taking \ L.C.M \big)\\ \\ \\ \dashrightarrow\sf \bigg[\dfrac{(4na-2na)+(4n^2d-n^2d)+(-2nd+nd)}{2}\bigg]\\ \\ \\ \dashrightarrow\sf \bigg[\dfrac{2na+3n^2d-nd}{2}\bigg]\\ \\ \\ \dashrightarrow\sf \bigg[\dfrac{ n(2a+3nd-d)}{2}\bigg]\ \ \ \ \ \ \big(taking\ 'n' \ as \ common \big)\\ \\ \\ \dashrightarrow\sf \dfrac{n[2a+(3n-1)d]}{2}\\ \\ \\ \dashrightarrow\sf \dfrac{n}{2}[2a+(3n-1)d]$

Now we have ,

$\dashrightarrow\sf S_2-S_1= \dfrac{n}{2}[2a+(3n-1)d]\\ \\ \\ \bullet\sf Multiplying \ both \ side \ by \ (3) \\ \\ \\ \dashrightarrow\sf 3\big(S_2-S_1\big)= 3\bigg(\dfrac{n}{2}[2a+(3n-1)d]\bigg)\\ \\ \\ \dashrightarrow{\red{\sf 3\big(S_2-S_1\big)= \dfrac{3n}{2}[2a+(3n-1)d]----- eq.\ (ii)}}$

★ From eq. (i) and eq.(ii) we can write -

$\underline{\boxed{\sf S_3=3(S_2-S_1)}}$

$\underline{\underline{\dag{\sf \ \ \ \ Hence \ Proved\ \ \ \ \ \ }}}$