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If the 10th term of  AP is 47 and its 1st term is 2.Find the sum of its first 15 terms

10th term of an APis given by;  a+9d = 47

and fir st term is given by; a = 2

now put the value of a in 1st eqn. to get , d= 5

now u have first term a, and common difference , d of the AP.

so sum of first 15 terms is given by;  S = 15/ 2 [ 2(2) + 14 (5) ] = 480. ans.

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a = 2

a+9d=47

ie d=5

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

ie 15/2[2*2+14*5]

thus sum of 15 terms is 480

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a10=a + 9d = 47 (given)  (   equtn 1 )

a=2 (given)

substituting value of a in equtn 1,we have

2 + 9d = 47

9d = 47-2

9d =45

implies that d = 45/9

=5

S15 =15/2(2a + (15-1)d )

S15 =15/2(2. 2 + 14.5) 

S15 = 15/2(4 + 70)

S15 = 15/2 .74

S15 = 15.37

S15 = 555

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ATQ >

a+9d = 47                   ------------------1

a = 2

putting a in 1

2+9d = 47

9d = 45

d = 5.

S15 = 15/2[2*2 + (15-1)5]

       = 15/2 [4 + 70]

       = 15 * 37

       = 555

thus, Jamesman is correct and rest are wrong.

hope this helps, plz give thumbs up

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thanks for thumbs up

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ua welcum frn......!!!!!!!

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ua welcum frn

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sorry to say that but ur ans is wrong

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480
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Answer please

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