Question

In an A.S, 6th term is 40 and 9th term is 58.Find the sum of 25 terms

Answer 1

Given,

a4+a6=42 …(1)

and

a3+a9=52 ...(2)

Using the formula for nth term,

an=a+(n−1)∗d

(1) becomes,

a+3d+a+5d=42

or 2a+8d=42

or a+4d=21 …..(a)

and (2) becomes,

a+2d+a+8d=52

or 2a+10d=52

or a+5d=26 …..(b)

Subtracting (a) from (b)we get,

d=5

and put this in (a), we get

a=1 .

Now, for the sum of first n terms, use the formula,

Sn=n2∗[2a+(n−1)d]

Therefore,

S10=5∗[2+(9)∗5]

S10=5∗47

S10=235

Answer 2

Answer:

a6 = 40

a+5d=40 (1)

a9= 58

a+8d=58 (2)

Subtracting eq (2) from (1)

a+8d=58

a+5d=40

-  -     -

    3d= 18

     d = 18/3

     d= 6

Substiting the value of d in eq (1)

a + 5(6) = 40

a+30=40

a=40-30

a=10

S25 = 25/2[2a+ (25-1)d]

      = 25/2[2*10+ 24*6]

      = 25/2 * 2 [10+12*6]      (2 is common inside the bracket)

      = 25 [10+ 72]

      = 25*82

S25 = 2050

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