# Arithmetic Progressions

#### Question 3:

In an AP

(i) Given *a* = 5,
*d* = 3, *a*_{n} = 50, find *n* and *S*_{n}.

(ii) Given *a* =
7, *a*_{13} = 35, find *d* and *S*_{13}.

(iii) Given *a*_{12}
= 37, *d* = 3, find *a* and *S*_{12}.

(iv) Given *a*_{3}
= 15, *S*_{10} = 125, find *d* and *a*_{10}.

(v) Given *d* = 5,
*S*_{9} = 75, find *a* and *a*_{9}.

(vi) Given *a* =
2, *d* = 8, *S*_{n} = 90, find *n* and
*a*_{n}.

(vii) Given *a* =
8, *a*_{n} = 62, *S*_{n} =
210, find *n* and *d*.

(viii) Given *a*_{n}
= 4, *d* = 2, *S*_{n} = − 14, find *n*
and *a*.

(ix) Given *a* =
3, *n* = 8, *S* = 192, find *d*.

(x)Given *l*
= 28, *S* = 144 and there are total 9 terms. Find *a*.

#### Answer:

(i) Given that, *a*
= 5, *d* = 3, *a*_{n} = 50

As
*a*_{n} = *a* + (*n* − 1)*d*,

∴
50 = 5 + (*n* − 1)3

45
= (*n* − 1)3

15
= *n* − 1

*n*
= 16

(ii) Given that, *a*
= 7, *a*_{13} = 35

As
*a*_{n} = *a* + (*n* − 1) *d*,

∴
*a*_{13} =* a* + (13 − 1) *d*

35
= 7 + 12 *d*

35
− 7 = 12*d*

28
= 12*d*

(iii)Given
that, *a*_{12} = 37, *d* = 3

As
*a*_{n} = *a* + (*n* − 1)*d*,

*a*_{12}
= *a* + (12 − 1)3

37
= *a* + 33

*a*
= 4

(iv) Given that, *a*_{3}
= 15, *S*_{10} = 125

As
*a*_{n} = *a* + (*n* − 1)*d*,

*a*_{3}
= *a* + (3 − 1)*d*

15
= *a* + 2*d* (i)

On multiplying equation (1) by 2, we obtain

30
= 2*a* + 4*d* (iii)

On subtracting equation (iii) from (ii), we obtain

−5
= 5*d*

*d*
= −1

From equation (i),

15
= *a* + 2(−1)

15
= *a* − 2

*a*
= 17

*a*_{10}
= *a* + (10 − 1)*d*

*a*_{10}
= 17 + (9) (−1)

*a*_{10}
= 17 − 9 = 8

(v)Given that,
*d* = 5, *S*_{9} = 75

As …

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