The $20^{\text {th }}$ term from the end of the progression $20,19 \frac{1}{4}, 18 \frac{1}{2}, 17 \frac{3}{4}, \ldots .,-129 \frac{1}{4}$ is :-

Insert five numbers between $8$ and $26$ such that resulting sequence is an $A.P.$

If the numbers $a,\;b,\;c,\;d,\;e$ form an $A.P.$, then the value of $a - 4b + 6c - 4d + e$ is

Consider a sequence whose sum of first $n$ -terms is given by $S_n = 4n^2 + 6n, n \in N$, then $T_{15}$ of this sequence is -

Let ${a_1},{a_2},{a_3}, \ldots $ be terms of $A.P.$  If $\frac{{{a_1} + {a_2} + \ldots + {a_p}}}{{{a_1} + {a_2} + \ldots + {a_q}}} = \frac{{{p^2}}}{{{q^2}}},p \ne q$ then $\frac{{{a_6}}}{{{a_{21}}}}$ equals

Which term of the sequence $( - 8 + 18i),\,( - 6 + 15i),$ $( - 4 + 12i)$ $,......$ is purely imaginary