JEE Main 2026 (Online) 2nd April Evening ShiftQuestion 1
Let $a_1,a_2,a_3,\ldots$ be an A.P. and $g_1=a_1,g_2,g_3,\ldots$ be an increasing G.P. If $a_1=a_2+g_2=1$ and $a_3+g_3=4$, then $a_{10}+g_5$ is equal to :
Sequence and Series · Mixed Progressions (AP-GP)
Solution
Formulas used. (1) $n$-th term of an A.P.: $a_n=a+(n-1)d$(2) $n$-th term of a G.P.: $g_n=ar^{n-1}$Step 1 - Set up. A.P.: first term $a_1=1$, common difference $d$. G.P.: $g_1=a_1=1$, common ratio $r$, so $g_2=r,\ g_3=r^2,\ g_5=r^4$. Step 2 - First condition $a_2+g_2=1$: $$(1+d)+r=1\ \Longrightarrow\ d=-r .$$ Step 3 - Second condition $a_3+g_3=4$: $$(1+2d)+r^2=4\ \Longrightarrow\ 1-2r+r^2=4\ \Longrightarrow\ (r-1)^2=4 .$$ So $r=3$ or $r=-1$. The G.P. is increasing, which rejects $r=-1$ (terms $1,-1,1,\dots$ alternate). Hence $r=3$ and $d=-3$. Step 4 - Required value. $$a_{10}=1+9(-3)=-26,\qquad g_5=3^4=81,$$ $$a_{10}+g_5=-26+81=55 .$$ Answer: (D)