Question

the sum of first 7 terms of an AP is 49 and that of first 17 terms of it is 289 find the sum of first n terms

Answer 1

Hi

Here devil1407

We know that
                    Sn = n/2 ( 2a + (n-1)d)
Sum of first 7 terms = 47
      
              S7 = 7/2 (2a + (n-1)d)
               49 = 7/2 (2a+(7-1)d)
               49 ×2/7 = 2a +6d
                   14 = 2a +6d
           ⇒  ( ÷ 2 )
                   a = 7 - 3d...............(1)

Sum of first 17 terms = 289

              S17 = 17/2 (2a+(n-1)d)
              289 = 17/2 (2a + (17-1)d)
           289 × 2/17 = 2a + 16d 
                 34 = 2a + 16d
        ⇒ ( ÷ 2 )  
                  a = 17 - 8d...........(2)

From (1) and (2) :

             7 - 3d = 17 - 8d
            - 3d +8d = 17 - 7
                   5d = 10
                    d = 10/5
                     d = 2 

Substitute d = 2 in equation (1)

                       a = 7 - 3d
                          = 7 - 3 × 2
                          = 7 - 6
                          = 1

 ∴ d = 2 and a =1..   

Sum of first n terms :

                Sn = n/2 (2 a + (n-1)d)
               Sn = n/2 ( 2 ×1 ( n-1) 2 )
                Sn  = n /2 ×2 (n)
∵ 2 GET CANCELED OUT.....
               Sn = n²

Hope it helps U

Thanx....Bye

Answer 2

$\sf{\bold{\green{\underline{\underline{Given}}}}}$

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• Sum of 7 terms of AP is 49
• Sum of 17 terms of AP is 289

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$\sf{\bold{\green{\underline{\underline{To\:Find}}}}}$

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• Sum of n terms = ??

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$\sf{\bold{\green{\underline{\underline{Solution}}}}}$

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Acc. to the 1st statement :-

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$\sf{\red{\boxed{\bold{S =\dfrac{n}{2} [ 2a + ( n - 1 ) d ] }}}}$

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$\sf :\implies\:{\bold{ 49 = \dfrac{7}{2}[ 2a + ( 7 - 1 ) d ]}}$

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$\sf :\implies\:{\bold{ 49 = \dfrac{7}{2}[ 2a + 6d ]}}$

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$\sf :\implies\:{\bold{ 49\times 2 = 7 [ 2a + 6d ]}}$

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$\sf :\implies\:{\bold{ \dfrac{49\times 2}{7} = [ 2a + 6d ]}}$

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$\sf :\implies\:{\bold{ 7\times 2 = [ 2a + 6 d ]}}$

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$\sf :\implies\:{\bold{ 7\times 2 = 2 [ a + 3 d ]}}$

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$\sf :\implies\:{\bold{ \dfrac { 7\times 2}{2} = [ a + 3 d ]}}$

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$\sf :\implies\:{\bold{ a + 3d = 7 }}$ --- ( i )

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Acc. to the 2nd statement :-

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$\sf{\red{\boxed{\bold{S =\dfrac{n}{2} [ 2a + ( n - 1 ) d ] }}}}$

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$\sf :\implies\:{\bold{ 289 = \dfrac{17}{2}[ 2a + ( 17 - 1 ) d ]}}$

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$\sf :\implies\:{\bold{ 289 = \dfrac{17}{2}[ 2a + 16d ]}}$

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$\sf :\implies\:{\bold{ 289\times 2 = 17 [ 2a + 16d ]}}$

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$\sf :\implies\:{\bold{ \dfrac{289\times 2}{17} = [ 2a + 16d ]}}$

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$\sf :\implies\:{\bold{ 17\times 2 = [ 2a + 16d ]}}$

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$\sf :\implies\:{\bold{ 17\times 2 = 2 [ a + 8d ]}}$

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$\sf :\implies\:{\bold{ \dfrac { 17\times 2}{2} = [ a + 8d ]}}$

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$\sf :\implies\:{\bold{ a + 8d = 17 }}$ ---- ( ii )

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• Subtracting eq ( i ) from ( ii )

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a + 8d - ( a + 3d ) =17 - 7

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a + 8d - a - 3d = 17 - 7

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8d - 3d = 17 - 7

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5d = 10

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d = 10/ 5

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d = 2

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$\sf{\underline{\boxed{\purple{\large{\bold{ d = 2 }}}}}}$

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• putting value of d in eq ( i )

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a + 3d = 7

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a + 3 × 2 = 7

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a + 6 = 7

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a = 7 - 6

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a = 1

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$\sf{\underline{\boxed{\purple{\large{\bold{ a = 1 }}}}}}$

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• Sum of n terms

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$\sf{\red{\boxed{\bold{S =\dfrac{n}{2} [ 2\times 1 + ( n - 1 ) 2 ] }}}}$

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$\sf :\implies\:{\bold{ S = \dfrac{n}{2}[ 2 + ( n - 1 ) 2 ]}}$

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$\sf :\implies\:{\bold{ S = \dfrac{n}{2}[ 2+ 2n - 2 ]}}$

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$\sf :\implies\:{\bold{ S = \dfrac{n}{2} [ 2n ]}}$

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$\sf :\implies\:{\bold{ S = n\times n}}$

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$\sf :\implies\:{\bold{ S = n^2}}$

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$\sf{\bold{\green{\underline{\underline{Answer}}}}}$

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• Sum of n terms = n²