Question

If 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

Answer:

Sn=n/2{-1510/13 + 282n/13 - 282/13}

Step-by-step explanation:

As per question

S7=7/2{2a+(7-1)d}

49=7/2{2a+6d}

49=14/2{a+3d}

a+3d=7............ .. ............i) equation

An=a+(17-1)d

289=a+16d

a+16d=289..........................ii) equation

by soving both equation

a+16d=289

a+3d=7

a= -755/13

d= 282/13

Therefore Sn= n/2[{2×(-755/13)} +(n-1)282/13]

Sn=n/2[{-1510/13} + 282n/13 - 282/13] answer

Answer 2

Given :-

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

• And 17 term of AP is 289

To find :-

• The sum of first n terms

Solution :-

we know that,

$\blue{\bigstar}\:\:{\underline{\boxed{\bf\red{s_n=\dfrac{n}{2}\:[2a+(n-1)d]}}}}$

$\rm\longrightarrow\:s_7=\dfrac{7}{2}\:[2a+(7-1)d]$

$\rm\longrightarrow\:49=\dfrac{7}{2}\:[2a+6d]$

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

$\rm\longrightarrow\:14=2a+6d$

$\rm\longrightarrow\:2a+6d=14...............(i)$

Again,

$\orange{\bigstar}\:\:{\underline{\boxed{\bf\pink{s_n=\dfrac{n}{2}\:[2a+(n-1)d]}}}}$

$\rm\longrightarrow\:s_{17}=\dfrac{17}{2}\:[2a+(17-1)d]$

$\rm\longrightarrow\:289=\dfrac{17}{2}\:[2a+16d]$

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

$\rm\longrightarrow\:34=2a+16d$

$\rm\longrightarrow\:2a+16d=34...............(ii)$

From equation (i) and (ii),

2a + 6d = 14

2a + 16d = 34

-⠀ -⠀ ⠀⠀ -

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-10d = -20

⤇ 10d = 20

⤇ d = 20/10

⤇ d = 2

Putting the value of d in equation (i),

2a + 6d = 14

⤇ 2a + 6 × 2 = 14

⤇ 2a + 12 = 14

⤇ 2a = 14 - 12

⤇ 2a = 2

⤇ a = 2/2

⤇ a = 1

Now, we have

• a = 1
• d = 2

$\pink{\bigstar}\:\:{\underline{\boxed{\bf\purple{s_n=\dfrac{n}{2}\:[2a+(n-1)d]}}}}$

$\rm\longrightarrow\:s_n=\dfrac{n}{2}\:[2\times{1}+(n-1)\times{2}$

$\rm\longrightarrow\:s_n=\dfrac{n}{2}\:[2+2n-2]$

$\rm\longrightarrow\:s_n=\dfrac{n}{2}\times{2n}$

$\rm\longrightarrow\:s_n=\dfrac{2{n}^{2}}{2}$

$\rm\longrightarrow\:s_n={n}^{2}$

Hence,the sum of first n terms will be n².