Question

give the sum of first 7 terms of an ap is 49 and that of 17 terms is 289 find the sum of first n terms​

Answer 1

Solution :

$\bigstar$ Firstly,we know that formula of the sum of an A.P;

$\boxed{\bf{S_n= \frac{n}{2}\bigg[2a+(n-1)d\bigg]}}}}$

• a is the first term.
• d is the common difference.
• n is the term of an A.P.

$\underline{\boldsymbol{According\:to\:the\:question\::}}}$

$\longrightarrow\sf{S_7 = 49}\\\\\longrightarrow\sf{\dfrac{7}{2} \bigg[2a+(7-1)d\bigg] = 49}\\\\\\\longrightarrow\sf{\dfrac{7}{2} \bigg[2a+6d\bigg] = 49}\\\\\\\longrightarrow\sf{14a + 42d = 98}\\\\\longrightarrow\sf{14(a + 3d) = 98}\\\\\longrightarrow\sf{a+3d =\cancel{98/14}}\\\\\longrightarrow\sf{a+ 3d = 7...................(1)}$

&

$\longrightarrow\sf{S_{17 }= 289}\\\\\longrightarrow\sf{\dfrac{17}{2} \bigg[2a+(17-1)d\bigg] = 289}\\\\\\\longrightarrow\sf{\dfrac{17}{2} \bigg[2a+16d\bigg] = 289}\\\\\\\longrightarrow\sf{34a + 272d = 578}\\\\\longrightarrow\sf{34(a + 8d) = 578}\\\\\longrightarrow\sf{a+8d =\cancel{578/34}}\\\\\longrightarrow\sf{a+ 8d = 17...................(2)}$

∴ Subtracting equation (1) & equation (2),we get;

$\longrightarrow\sf{\cancel{a-a} + 3d - 8d = 7 -17 }\\\\\longrightarrow\sf{-5d = -10}\\\\\longrightarrow\sf{d =\cancel{-10/-5}}\\\\\longrightarrow\bf{d = 2}$

Putting the value of d in equation (1),we get;

$\longrightarrow\sf{a + 3(2) = 7}\\\\\longrightarrow\sf{a + 6 =7}\\\\\longrightarrow\sf{a = 7-6}\\\\\longrightarrow\bf{a = 1}$

Now;

$\longrightarrow\sf{S_n=\dfrac{n}{2} \bigg[2(1) + (n-1)(2)\bigg]}\\\\\\\longrightarrow\sf{S_n= \dfrac{n}{2} \bigg[\cancel{2} + 2n \cancel{- 2} \bigg]}\\\\\\\longrightarrow\sf{S_n = \dfrac{n}{\cancel{2}} \times \cancel{2}n}\\\\\longrightarrow\sf{ S_n = n\times n}\\\\\longrightarrow\bf{S_n = n^{2}}$

Thus;

The sum of first n terms will be n² .

Answer 2

$\large\sf\underline\red{GIVEN:-}$

$\large\tt\pink{S7=49}$

$\large\tt\pink{S17=289}$

$\small\sf\underline\blue{ The\:sum\:of\:n\: terms\:of\: an \:AP\: is \:given \:by}$

$\longrightarrow$$\large\tt\purple{Sn=\frac{n}{2}(2a+(n-1)d)}$

$\longrightarrow$$\large\tt\purple{S7=\frac{7}{2}(2a+(7-1)d)}$

$\longrightarrow$$\large\tt\purple{49=\frac{7}{2}(2a+6d)}$

$\longrightarrow$$\large\tt\purple{49=7(a+3d)}$

$\longrightarrow$$\large\tt\purple{7=7-3d.......(1)}$

$\huge\sf\underline\red{And,}$

$\longrightarrow$$\large\tt\purple{S14=\frac{17}{2}(2a+(17-1)d)}$

$\longrightarrow$$\large\tt\purple{289=\frac{17}{2}(2a+16d)}$

$\longrightarrow$$\large\tt\purple{289=17(a+8d)}$

$\longrightarrow$$\large\tt\purple{17=a+8d}$

$\small\sf\underline\pink{Putting\:the\:value\:of\:a\:from\:eq(1),we\:get}$

$\longrightarrow$$\large\tt\purple{17=7-3d+8d}$

$\longrightarrow$$\large\tt\purple{5d=10}$

$\longrightarrow$$\large\tt\purple{d=2}$

$\small\sf\underline\pink{Putting\:the\:value\:of\:d\:from\:eq(1),we\:get}$

$\longrightarrow$$\large\tt\purple{a=7-3×2}$

$\longrightarrow$$\large\tt\purple{a=1}$

$\small\sf\underline\pink{The\:sum\:of\:n\:terms\:of\:an\:AP\:is\:given\:by}$

$\longrightarrow$$\large\tt\purple{Sn=\frac{n}{2}(2a+(n-1)d)}$

$\longrightarrow$$\large\tt\purple{\frac{n}{2}(2(1)+(n-1)(2))}$

$\longrightarrow$$\large\tt\purple{\frac{n}{2}(2+2n-2}$

$\longrightarrow$$\large{\boxed{\tt{\purple{{n}^{2}}}}}$