Question

the 10th term and the 18th term of an A.P.are 25 and 41 respectively . find 1) the 1st term and the common difference 2) the 34th term 3) n such that tn= 87
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Answer 1

Step-by-step explanation:

Let a is first term and d is common difference of an A.P.

tn=a+(n-1)d

t10=a+(10-1)d

25=a+9d

a+9d=25..... (1)

tn=a+(n-1)d

t18=a+(18-1)d

41=a+17d

a+17d=41.... (2)

Subtracting equation(1)from(2)

a+9d=25

- a+17d=41

- - -

----------------------

-8d= -16

d= -16/-8

d= 2

Put d=2 in equation (1)

a+9d=25

a+9(2)=25

a+18=25

a=25-18

a=7

1) First term=a

=7

Common difference=d=2

2) tn=a+(n-1)d

t34=7+(34-1)2

t34=7+35×2

t34=7+70

t34=77

The 34th term is 77.

3) tn=a+(n-1)d

87=7+(n-1)2

87-7=2n-2

80=2n-2

80+2=2n

82=2n

n=82/2

n=41

Answer 2

Step-by-step explanation:

T10=25

T18=41

1)the 1st term and the common difference

a=?,d=?

d=T18-T10

18-10 T18=41

a+17d=41

=41-25 a+17×2=41 . 8 a + 34 = 41

a=41-34

=16 a=7

8

d=2

2)the34th term

T34=?

Tn =a+(n-1)d

T34=7+(34-1)2

T34= 7+33×2

T34=7×66

T34=5082

3)n such that Tn=87

Tn=a+(n-1)d

87=7+(n-1)2

87=7+2n-2

87=5+2n

87-5=2n

82=2n

n=82

2

n=41

T41=82