Question

slove it
1. In the AP 12, 19, 26..... find a15
2. In the AP 7, 13, 19, ....... 151 find number of term
3. If 2 n is equal to 2 n square + 5 find 12th term (a12)
4. If an=n2/n+1, find first four term (a,a2, a3, a4)
5. If 8th term of AP is 17 and 19th term is 39, then find 25th term
6. if 4th term is 17 and 10th term is exceeds the 7th term by 12 find first 3 terms of AP
7. the sum of 4th and 8th term of AP is 24 and sum of 6th and 10th term is 44 find a and d
8. if 11 times the 11th term= 7 times the 7th term s.t 18th term of AP​

Answer 1

Step-by-step explanation:

dear for such a long and hard question u have to give some more points

Answer 2

1.a1=12,a2=19,a3=26,a15=?.

a15 = a+14d = 12+14(7) = 110. {•d= common difference}.

2.a1=7,a2=13,a3=19,a?=151.

an= a+(n-1)d

151=7+(n-1)6 = 151-7 =(n-1)6 = 144/6 =(n-1) = 24+1=n.

(•n= 25).

3.I can't understand it.

4.In case an=n square/n+1.

• if n=1, 1/2.
• if n=2,4/3.
• if n=3,9/4.
• if n=4,16/5.(•a1=1/2,a2=4/3,a3=9/4,a4=16/5).

5.the 8th term of AP is 17, 19th term is 39.

1. 18th term = a+7d= 17.
2. 19th term =a+18d= 39.

(1-2) = (elimination method) -11d= -22,=d=22/11,d=2.

from1:- a+7(2)=17, a+14=17,a=17-14=3.

(• a=3,d=2).

then 25th term is a+24d.

● 3+24(2),3+48=51

●Therefore 25th term is 51.

6.Given that 4th term of AP =17.

therefore a+3d=17.

also given that 10th term exceeds the 7th by 12.

therefore a10 = a7+12,a+9d = a+6d +12.

• a+9d-a-6d = 12
• 3d =12
• d = 12/3
• d = 4.

From1 a+3(4) = 17, a+12 =17,a = 17-12, a=5.

Therefore first three terms are 5,9,13.

7.the sum of 4th term & 8th term

is 24, the sum of 6th term & 10th term is 44.so, find a &d.

1.a+3d+a+7d = 24.

2.a+5d+a+9d = 44.

• 1. 2a+10d = 24, 2. 2a+14d =44.

(1-2) that implies by (elimination method) -4d = -20

d = 20/4 (● d = 5)

from1:- 2a+10(5) =24, 2a = 24-50, 2a = -26

a = -26/2, a = -13.

Therefore a = -13, d = 5.

8. I cannot understand it please post it clearly.

I hope it is helpful to you.