- A$5, 9, 11, 13$
- ✓$7, 11, 15, 19$
- C$5, 11, 15, 22$
- D$7, 15, 19, 21$
So $3,\;{A_1},\;{A_2},\;{A_3},\;{A_4},\;23$
$ \Rightarrow $ ${T_6} = 23 = a + 5d$
$ \Rightarrow $ $d = 4$
Thus ${A_1} = 3 + 4 = 7,\;{A_2} = 7 + 4 = 11,\;$
$A_3 = 11+4 =15 ,\, A_4 = 15+ 4=19$
Generate a complete, print-ready paper with questions like this in minutes — across 16+ boards, with answer keys.
$T_{\mathrm{I}}=V_{\mathrm{r}+1}-V_{\mathrm{I}}-2 \text { and } \mathrm{Q}_{\mathrm{I}}=T_{\mathrm{r}+1}-\mathrm{T}_{\mathrm{r}} \text { for } \mathrm{r}=1,2, \ldots$
$1.$ The sum $V_1+V_2+\ldots+V_n$ is
$(A)$ $\frac{1}{12} n(n+1)\left(3 n^2-n+1\right)$
$(B)$ $\frac{1}{12} n(n+1)\left(3 n^2+n+2\right)$
$(C)$ $\frac{1}{2} n\left(2 n^2-n+1\right)$
$(D)$ $\frac{1}{3}\left(2 n^3-2 n+3\right)$
$2.$ $\mathrm{T}_{\mathrm{T}}$ is always
$(A)$ an odd number $(B)$ an even number
$(C)$ a prime number $(D)$ a composite number
$3.$ Which one of the following is a correct statement?
$(A)$ $Q_1, Q_2, Q_3, \ldots$ are in $A.P.$ with common difference $5$
$(B)$ $\mathrm{Q}_1, \mathrm{Q}_2, \mathrm{Q}_3, \ldots$ are in $A.P.$ with common difference $6$
$(C)$ $\mathrm{Q}_1, \mathrm{Q}_2, \mathrm{Q}_3, \ldots$ are in $A.P.$ with common difference $11$
$(D)$ $Q_1=Q_2=Q_3=\ldots$
Give the answer question $1,2$ and $3.$