Question

Plz answer these questions....

I want answers if all questions plz.....

Answer 1

to find 11th tern from last.. just reverse ap so that it becomes 11th one from beginning
ap becomes
-62,......4,7,10
a=-62
d=10-7=3
11th tern =a+10d=-62+10(3)=-32

(2).
$a + 2d = 16$
$a + 4d + 12 = a + 6d$
solve
4d+12=6d

d=6
put in first eqn
$a + 2(2) = 16 \\ a = 16 - 4 \\ a = 12$
ap is
a, a+d, a+2d,......
12,18,24,......

(3).
1+4+7+10+....+x=287
let the number of terms be n
sum of Lhs
=(n/2)(2(1)+(n-1)(3))=287
=n(2+3n-3)=287*2
=n(3n-1)=574
3n^2-n-574=0
solve and take positive value of n
n=14
14th term of this ap
x=a+13d
x=1+13(3)=40

Answer 2

4) According to given sum,

10,7,4.......-62

a=10, d=7-10=-3

an=a+(n-1)d

-62=10+(n-1)(-3)

(n-1)(-3)=-72

n-1=-72/-3=24

n=25 .

11th term from last is equal to the( (25-11)+1 )th term from starting .

so 15th term from first .

a15=a+(n-1)d

=10+14 (-3)

=-32 .

5) According to given sum,

a+2d=16............(i)

seventh term exceeds 5th term by 12 .

a+6d=a+4d+12

6d-4d=12

2d=12

d=6.

substitute the d value in equation i.

a+2 (6)=16

a=16-12=4

The AP is in the form of

a,a+d,a+2d..................

4,4+6,4+2 (6).................

4,10,16............

6) 1+4+7+10............+x=287

a=1, d=4-1=3

sn=n/2 (2a+(n-1)d)

287=n/2 (2*1+(n-1)3)

287×2=n (2+3n-3)=3n^2-n .

574=3n^2-n

3n^2-n-574=0

3n^2-42n+41n-574=0

3n (n-14)+41 (n-14)=0

(n-14)(3n+41)=0

n=14 or -41/3

n always be a natural number it won't be any fraction. so n=14

an=a+(n-1)d

x=1+(14-1)3

=1+39

=40 .

so x=40 .

:-)hope it helps u.

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