Ratios with A.P.

Question

Two A.P.’s have the same number of terms. The ratio of the last term of the first progression to first term of the second progression is equal to the ratio of the last term of the second progression to the first term of the first progression and is equal to 4, the ratio of the sum of the n-terms of the first progression to the sum of the n-terms of the second progression is equal to 2.

1) What is the ratio of Common Difference?

2) What is the ratio of n-th term?

3) What is the ratio of first term?

ANSWER:

n₁=n₂

t_n1/a₂ = t_n2/a₁ = 4

S_n1/S_n2 = 2

Since A.P. deals with addition, I don't know how to cancel the unnecessary terms in the ratio.

Any help would be greatly appreciated!!!

Answer 1

HINT: if $$\frac{a}{b}=\frac{c}{d}$$ then each is equal to $\frac{a+c}{b+d}$ and $\frac{a-c}{b-d}$

Answer 2

See if two ratios are given then we can use two rules componendo ,dividendo so here we write $a_n=a+(n-1)d$ ,$a'_n=a'+(n-1)d'$ so we want $d/d',\frac{a+a+(n-1)d}{a'+a'+(n-1)d'}$ so first one is $\frac{d}{d'}=\frac{a+(n-1)d-a}{a'+(n-1)d'-a'}=d/d'=4$ same logic goes for other