Question

answer these questions guys ​

Answer 1

Answer:

Q32. 100

Q33. n= 13

Q34. Use the same method as the last question

Q35. Proven in the explanation

Step-by-step explanation:

Q32.

In AP1, a= first term , d = common difference ,

In AP2 , a' = first term , d= common difference

100th term of AP1= a + 99d

100th term of AP2 = a' + 99d (because nth term = a +(n-1)d)

And, a + 99d -(a' + 99d) = 100

        a- a' = 100

Difference between 1000th terms

a + 999d - ( a' + 999d ) = a - a' =100

Q33.

In AP1, nth term = 63 + (n-1)2 = 61 + 2n

In AP2 , nth term = 3 + (n-1)7 = 7n - 4

As they are equal, 61 +2n = 7n - 4

therefore, n=13

Q34. Use the same method as before

Q35.

8th term = a +7d =0

hence, a= -7d

38th term =  a +( 38-1)d = -7d + 37d = 30d

18th term = a +(18 - 1)d = -7d +17d = 10d

Hence proven that 38th term = 3* 18th term