Insert $6$ numbers between $3$ and $24$ such that the resulting sequence is an $A.P.$
Let $A_{1}, A_{2}, A_{3}, A_{4}, A_{5}$ and $A_{6}$ be six numbers between $3$ and $24$ such that $3, A _{1}, A _{2}, A _{3}, A _{4}, A _{5}, A _{6}, 24$ are in $A.P.$ Here, $a=3, b=24, n=8$
Therefore, $24=3+(8-1) d,$ so that $d=3$
Thus ${A_1} = a + d = 3 + 3 = 6;\quad $
${A_2} = a + 2d = 3 + 2 \times 3 = 9$
${A_3} = a + 3d = 3 + 3 \times 3 = 12;\quad $
${A_4} = a + 4d = 3 + 4 \times 3 = 15$
${A_5} = a + 5d = 3 + 5 \times 3 = 18;\quad $
${A_6} = a + 6d = 3 + 6 \times 3 = 21$
Hence, six numbers between $3$ and $24$ are $6,9,12,15,18$ and $21$
If all interior angle of quadrilateral are in $AP$ . If common difference is $10^o$ , then find smallest angle ?.....$^o$
If $a,b,c$ are in $A.P.$, then $\frac{1}{{\sqrt a + \sqrt b }},\,\frac{1}{{\sqrt a + \sqrt c }},$ $\frac{1}{{\sqrt b + \sqrt c }}$ are in
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
Let $S_n$ denote the sum of the first $n$ terms of an arithmetic progression. If $\mathrm{S}_{10}=390$ and the ratio of the tenth and the fifth terms is $15: 7$, then $S_{15}-S_5$ is equal to:
The sum of first $n$ natural numbers is