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How many terms of the Arithmetic Progressions 78,71,64,.........are needed for the sum to give the sum 465? Also find the last term of the AP?

let the required terms be n,

now a/q,

a=78 

d=71-78 =-7

sn=465

now 2s=n{2a+(n-1)d}

putting the values of s,a and d we get,

2*465=n{2*78+(n-1)-7}

930=163n-7n2

7n2-163n+930=0

On solving the above equation we will get get the required value n.

now ,

let the last term be l

Then,

l=a+(n-1)d

on putting the value of a,n and d we will get the last term.

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If you got then

THUMBZZZZ UP PEASZZZZ

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 Sn = 465 , a = 78 , d = 7

  AP = 78,71,64...........

  Sn = n/2 [2a + (n - 1) d]

  465= n/2 [2 x 78 + 7n - 7]

  930 = 156n +7n2 - 7n

  930 = 149n + 7n2

 =  7n2 + 149n - 930

    Factorize this eq to get n i.e,

    the no of terms req to give the sum as 465

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 Given: Sn=465

Sn=(n/2)(2a+(n-1)d)

ð  2Sn=n(2a+(n-1)d)

ð  2(465)=n(2(78)+(n-1)(-7))

ð  930=n(156+7-7n)

ð  930=n(163-7n)

ð  930=163n-7n2

ð  7n2-163n+930=0

ð  7n2-70n-93n+930=0

ð  7n(n-10)-93(n-10)=0

ð  n=10;(93/7)

(93/7) cannot be a value for ‘n’ as the number of terms (n) can never be a fraction.

Therefore, n=10

Last term=l

l=a10

an=a+(n-1)d

ð  a10=78+(10-1)(-7)

ð  a10=78+9(-7)

ð  a10=78+(-63)

ð  a10=78-63

ð  a10=15

ð  l=15

Therefore, last term is 15 

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