Question

If the 20th term of an AP is five times its third term . If its 10th term is 41, then find the sum of its first fifteen term

Answer 1

To find first 15 term in AP
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It is given,

20th term of the AP, A(20)=5 x 3rd term(A3)

So,

By A (n) = a + (n-1)d
A(20) = a + (20-1)d

A(20) = a + (19)d------------(1)

A(3) = a + (2)d-------------(2)

And, For A(10) =41

A(10) = a + (10-1)d--------(3)
A(10) = a + (9)d---------(3)
41 = a+ 9d ------------(3)
a= 41 - 9d-----------(3)

So by the given situation,

20th term of the AP, A(20)=5 x 3rd term(A3)

A(20) = 5 (A3)

a + (19)d= 5× {a + (2)d}
a + 19d= 5a + 10d
a-5a= 10d - 19d
-4a= -9d
4a= 9d

As, a= 41 - 9d-----------(3)

So,

4× (41 - 9d) = 9d
164 - 36d = 9d
164= 9d+36d
164= 45 d
d= 164/45

And putting value of D in a question 3rd.

a= 41 - 9d
a= 41 - 9 (164/45 )
a= 41 - 164/5
a = (205 - 164)/5
a= 41/5

Now,

Sum of first 15 terms

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

S (15) = 15/2 { 2(41/5)+ (15-1)(164/45)}

S (15) = 15/2 { (81)/5+ (14×164)/45)}

S (15) =15/2 (738 + 2296)/45

S (15) = 15/2× 3034/45

S (15) = 1517/3

is the answer

Answer 2

Third term = a + 2d
20th term = a + 19

×××××××××××××××××××
According to the given information,

5(a + 2d) = a + 19d

=> 5a + 10d = a + 19d

=> 5a - a = 19d - 10d

=> 4a = 9d

=> a = 9/4 d ------1equation

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10th term = a + 9d

41 = a + 9d

Putting the value of a from 1equation

41 = 9/4 d + 9d

=> 41 =(9 + 36)/4 ×d

=> 41 = 45/4 × d

=> 164/45 = d

×××××××××××××××××××

Putting the value of d in 1equation,

a = 9/4 × 164/45

a = 41/5

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Main content:-

15th term= a + 14d

=> 41/5 + 14(164/45)

=> 41/5 + 2296/45

=> (369+2296)/45

=> 2665/45

=> 533/9

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S(n) = (a + l) × n/2

=> S(15) = [ a + a(15)] × 15/2

=> S(15) = [ 41/5 + 2665/45] × 15/2

S(15) = [ 3034/45] × 15/2

S(15) = 1517/3

Sum of 15 terms = 1517/3

I hope this will help you

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