Question

If the sum of first 7 terms of AP is 49 and that of 17 terms is 289, find the sum of first n terms

Answer 1

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

• The sum of first 7 terms of an AP is 49

• And the sum of first 17 terms is 289.

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

• The sum of first n terms.

$\large\bf{\underline{\underline\green{Solution:-}}}$

As we know that:-

Sum of n terms is given by the formula:-

$\boxed{ \rm \leadsto S_{n} = \frac{n}{2} \bigg \{2a + (n - 1)d \bigg \} }$

Sum of first 7 terms:-

• n = 7

• $\rm S_{7} = 49$

$\rm \implies S_{7} = \dfrac{7}{2} \bigg \{2a + (7 - 1)d \bigg \}$

$\rm \implies 49= \dfrac{7}{2} \bigg \{2a + 6d \bigg \}$

$\rm \implies 49 \times \dfrac{2}{7} = 2a + 6d$

$\rm \implies 7 \times 2 = 2a + 6d$

$\rm \implies 2a + 6d = 14$

$\rm \implies a + 3d = 7.......(i)$

Sum of 17 terms:-

• n = 17

• $\rm S_{17} = 289$

$\rm \implies S_{17} = \dfrac{17}{2} \bigg \{2a + (17 - 1)d \bigg \}$

$\rm \implies 289= \dfrac{17}{2} \bigg \{2a + 16d \bigg \}$

$\rm \implies 289 \times \dfrac{2}{17} = 2a + 16d$

$\rm \implies 17 \times 2 = 2a + 16d$

$\rm \implies 2a + 16d = 34$

$\rm \implies a + 8d = 17.......(ii)$

Adding equation (i) and (ii)

$\rm \: \: \: \not a + 8d = 17$

$\rm - \not a - 3d = -7$
________________
$\rm \implies 8d - 3d = 17 - 7$

$\rm \implies 5d = 10$

$\rm \implies d = 2$

Substituting d = 2 in equation (i)

$\rm \implies a + 3 d = 7$

$\rm \implies a + 3 (2) = 7$

$\rm \implies a + 6 = 7$

$\rm \implies a = 1$

We have :- d = 2 and a = 1

Sum of n terms

${ \rm \leadsto S_{n} = \frac{n}{2} \bigg \{2a + (n - 1)d \bigg \} }$

$\rm \implies S_{n} = \dfrac {n}{2} \bigg \{2 \times 1 + (n - 1)2 \bigg \}$

$\rm \implies S_{n} = \dfrac {n}{2} \bigg \{2 + 2n - 2 \bigg \}$

$\rm \implies S_{n} = \dfrac { {2n}^{2} }{2}$

$\rm \implies S_{n} = { {n}^{2} }$

$\large \boxed{\rm \therefore Sum \: of \: n \: terms \:= { {n}^{2} } }$

Answer 2

$\bf\huge\blue{\underline{\underline{ Question : }}}$

If the sum of first 7 terms of AP is 49 and that of 17 terms is 289, find the sum of first n terms.

$\bf\huge\blue{\underline{\underline{ Solution : }}}$

★ Given that,

$\tt \rightarrow Sum\:of\:First\:7\:terms\:of\:AP = 49$

• $\tt S_{7} = 49$

$\tt \rightarrow Sum\:of\:First\:17\:terms\:of\:AP = 289$

• $\tt S_{17} = 289$

★ To find,

$\tt \rightarrow Sum\:of\:First\:n\:terms\:of\:an\:AP.$

★ Formula :

Sum of first n terms of an AP is :

$\boxed{\rm{\red{ S_{n} = \cfrac{n}{2} \Bigg [ 2a + (n-1)d \Bigg] }}}$

★ Let,

◼ ᴄᴀsᴇ - 1 :-

• S7 = 49
• n = 7

$\sf \implies 49 = \cfrac{7}{2} \Bigg [ 2a + (7-1)d \Bigg ]$

$\sf \implies \cancel{49} \times \cfrac{2}{\cancel{7}} = 2a + 6d$

$\sf \implies 7 \times 2 = 2a + 6d$

$\sf \implies 2a + 6d = 14$

$\sf \implies 2(a + 3d)=14$

$\sf \implies a + 3d= \cancel{\cfrac{14}{2}}$

$\sf \implies a + 3d = 7 .....(1)$

◼ ᴄᴀsᴇ - 2 :-

• S17 = 289
• n = 17

$\sf \implies 289 = \cfrac{17}{2} \Bigg[ 2a + (17 - 1)d \Bigg]$

$\sf \implies \cancel{289} \times \cfrac{2}{\cancel{17}} = 2a + 16d$

$\sf \implies 17 \times 2= 2a + 16d$

$\sf \implies 2a + 16d = 34$

$\sf \implies 2(a + 8d) = 34$

$\sf \implies a + 8d= \cancel{\cfrac{34}{2}}$

$\sf \implies a+8d = 17 ..... (2)$

• Subtract (1) & (2). We get,

$\sf \implies - 5d = -10$

$\sf \implies d =\cancel{\cfrac{ -10}{-5}}$

$\sf \implies d = 2$

• Substitute the value of d in (1).

$\sf \implies a + 3(2)=7$

$\sf \implies a + 6=7$

$\sf \implies a = 7-6$

$\sf \implies a = 1$

★ Verification,

• Substitute the values of a & d in (1), to get LHS = RHS.

LHS =

$\sf \leadsto 1 + 3(2)$

$\sf \leadsto 1 + 6$

$\sf \leadsto 7$

Since,LHS = RHS. Hence,it was verified.

★ Now,

We have the terms :-

• a = 1
• d = 2

Now, we can find out the sum of n terms of an AP.

$\sf \implies S_{n} = \cfrac{n}{2} \Bigg[ 2(1)+(n-1)2 \Bigg]$

$\sf \implies S_{n} = \cfrac{n}{2} \Bigg[ 2+2n-2 \Bigg]$

$\sf \implies S_{n} = \cfrac{n}{2} \Bigg[ 2n \Bigg]$

$\sf \implies S_{n} = \cfrac{\cancel{2}n^{2}}{\cancel{2}}$

$\sf \implies S_{n} = n^{2}$

$\underline{\boxed{\rm{\purple{\therefore Sum\:of\:n\:terms\:of\:an\:AP\:is\:n^{2}.}}}}\:\orange{\bigstar}$

★ More Information,

$\boxed{\begin{minipage}{5 cm} AP Formulae \\ \\ : \implies a_{n} = a + (n - 1)d \\ \\ :\implies S_{n} = \frac{n}{2} [ 2a + (n - 1)d ] \end{minipage}}$

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