Suppose that all the terms of an arithmetic progression ($A.P.$) are natural numbers. If the ratio of the sum of the first seven terms to the sum of the first eleven terms is $6: 11$ and the seventh term lies in between $130$ and $140$ , then the common difference of this $A.P.$ is

If $a_1, a_2, a_3, .... a_{21}$ are in $A.P.$ and $a_3 + a_5 + a_{11}+a_{17} + a_{19} = 10$ then the value of $\sum\limits_{r = 1}^{21} {{a_r}} $ is 

The number of terms common between the two series $2 + 5 + 8 +.....$ upto $50$ terms and the series $3 + 5 + 7 + 9.....$ upto $60$ terms, is

The number of common terms in the progressions $4,9,14,19, \ldots \ldots$, up to $25^{\text {th }}$ term and $3,6,9,12$, up to $37^{\text {th }}$ term is :

The sums of $n$ terms of two arithmetic progressions are in the ratio $5 n+4: 9 n+6 .$ Find the ratio of their $18^{\text {th }}$ terms.

Let $a_1, a_2 , a_3,.....$ be an $A.P$, such that $\frac{{{a_1} + {a_2} + .... + {a_p}}}{{{a_1} + {a_2} + {a_3} + ..... + {a_q}}} = \frac{{{p^3}}}{{{q^3}}};p \ne q$. Then $\frac{{{a_6}}}{{{a_{21}}}}$ is equal to