Question

In an AP :
1) Given a= 5, d= 3, an=50, find n and Sn
2) Given a=7, a13 = 35, find d and S13
3) Given a12= 37, d= 3 , find a and S12
4) Given a3= 15, S10 = 125, find d and a10
5) Given d= 5, S9= 75, find a and a9
6) Given a=2, d=8,Sn= 90, find n and an
7) Given a=8, an= 62, Sn= 210, find n and d
8) Given an= 4, d=2, Sn= -14, find n and a
9) Given a=3, n=8,S=192, find d
10) Given l = 28, S = 144, and there are total 9 terms. Find a

Answer 1

Hey I'm MrMysterious

1) We have, a= 5, d=3 and an=50

> a+(n-1)d= 50

> 5+(n-1)3=50

> 3(n-1)=50-5

> n-1 = 45/3

> n-1 = 15+1 = 16

Putting n = 16, a= 5 and l = an = 50 in Sn= n/2(a+l),
We get

S16 = 16/2( 5 + 50) = 8 × 55 = 440

Hence, n= 16 and S16 = 440

2) We have , a= 7 and a13 = 35

Let d be the common difference of the given AP. Then,

> a13 = 35

> a+ 12d = 35

> 7 + 12d = 35 [ °.° a = 7 ]

> 12d = 35 - 7

> 12d = 28

> d = 28/12

> d = 7/3

Putting n = 13, a = 7 and l = a13 = 35 in

> Sn = n/2(a+l), we get

> S13= 13/2(7+35)

> S13= 13/2×42

> S13= 13×21 = 273

Hence, d = 7/3 and S13 = 273

3) We have , a12= 37, d = 3

Let a be the first term of the given AP. Then,

> a12 = 37

> a + 11d = 37

> a + 11(3) = 37 [ °.° d = 3 ]

> a = 37-33

> a = 4

Putting n= 12, a= 4 and l = a12 = 37 in

> Sn = n/2(a+l), we get

> S12 = 12/2(4+37)

> 6× 41 = 246

Hence, a= 4 and S12 = 246

4) We have , a3 = 15, S10= 125

Let a be the first term and d the common difference of the given. Then,

> a3 = 15 and S10= 125

> a + 2d = 15 ........(1)

> and, 10/2[ 2a + (10 - 1)d] = 125

> 5(2a+9d) = 125

> 2a+ 9d = 25 ........(2)

> 2×(1) - (2) gives,

> 2(a + 2d) - (2a + 9d) = 2× 15 - 25

> 4d- 9d = 30-25

> -5d = 5

> d = -5/5

> d = -1

> Now, a10 = a+ 9d = ( a + 2d) + 7d

> 15+ 7(-1) [ Using (1) ]

> 15-7 = 8

Hence, d = -1 and a10 = 8

5) We have , d = 5 , S9 = 75

Let a be the first term of the given AP. Then,

> S9 = 75

> 9/2[2a + (9-1)5] = 75

> 9/2(2a+40) = 75

> 9a + 180 = 75

> 9a = 75 + 180

> 9a = -105

> a = -105/9 = -35/3

Now, a9 = a + 8d = -35/3+ 8 × 5

> -35+120/3 = 85/3

Hence , a = -35/3 and a9 = 85/3

5) We have , a= 2, d = 8, Sn= 90

> Sn = 90

> n/2[2×2+(n-1)8] = 90

> n/2( 4 + 8n - 8) = 90

> n/2(8n - 4) = 90

> n(4n - 2) = 90

> 4n²- 2n - 90 = 0

> .°. n = -(-2)√(-2)²-4×4×(-90)/ 2× 4

> 2±√4+1440/8

> 2±√1444/8

> 2±38/8

> 40/8, -36/8

> 5,-9/2

But n cannot be negative
> .°. n= 5

> Now , an = a+ (n-1)d

> a5 = 2 + (5-1)8

> 2+32

> 34

Hence , n= 5 and an= 34

7) We have , a = 8, an = 62 , Sn = 210

> Let d be the common difference of the given AP

Now, Sn = 210
> n/2(a+l)= 210

> n/2(8+62) = 210 [ °.° a = 8, an = 62]

> n/2×70 = 210

> n= 210 × 2/70 = 3 × 2 = 6

and , an= 62 a6= 62

> a+ 5d = 62

> 8+ 5d = 62

> 5d = 62-8

> d = 54/5

Hence, d = 54/5 and n = 6

8) We have , an = 4 , d = 2 , Sn = -14

Let a be the first term of the given AP. Then,

> an = 4

> a+ (n-1)2 = 4

> a = 4-2(n-1) ........(1)

> and , Sn = -14

> n/2(a+l) = -14

> n(a+l) = -28

> n[4-2(n-1)+4] = -28

> n( 4 - 2n + 2+4) = -28

> n( -2n + 10) = -28

> n(-n+5) = -14

> -n²+5n = -14

> n²- 5n - 14 = 0

> (n-7)(n+2) = 0

> n = 7 or -2

But n cannot be negative

> n = 7

Putting n = 7 in (1), we get

> a = 4-2( 7 -1) = 4 - 2 × 6

> = 4-12

> = -8

Hence, n = 7 and a = -8

9) We have, a = 3, n = 8, S = 192

Let d be the common difference of the given AP,

> Sn = n/2[2a + (n-1)d]

> 192 = 8/2 [ 2×3+(8-1)d]

> 192 = 4(6+7d)

> 48 = 6 + 7d

> 7d = 48-6

> 7d = 42

> d = 42/7 = 6

Hence, d = 6

10) We have,l = 28, S = 144, n= 9

Let a be the first term of the given AP

> S = 144

> n/2(a+l) = 144

> 9/2(a+ 28) = 144

> a + 28 = 144× 2/9

> a + 28 = 32

> a = 32-28

> a = 4

Hence , a = 4



?MrMysterious?

Answer 2

Answer:

I hope it is helpful....................