JEE Main 2025 Jan 22 Shift 2, Mathematics Q5: Arithmetic, Geometric And Harmonic Progressions

Let the A.P. have first term $a$ and common difference $d$, with $2k$ terms. The last term is $a+(2k-1)d$.

Last $-$ first $=(2k-1)d=27\Rightarrow d=\frac{27}{2k-1}$ .... (i)

Odd-position terms ($a_1,a_3,\dots ,a_{2k-1}$) form an A.P. of $k$ terms with first term $a$ and common difference $2d$. Their sum:

$S_{\text{odd}}=\frac{k}{2}\left[2a+(k-1)\cdot 2d\right]=k\left[a+(k-1)d\right]=40$.

Even-position terms similarly form an A.P. of $k$ terms with first term $a+d$:

$S_{\text{even}}=k\left[a+d+(k-1)d\right]=k\left[a+kd\right]=55$.

Subtracting: $S_{\text{even}}-S_{\text{odd}}=k\cdot d=55-40=15$, so $d=\frac{15}{k}$ .... (ii)

Equating (i) and (ii): $\frac{27}{2k-1}=\frac{15}{k}\Rightarrow 27k=15(2k-1)=30k-15\Rightarrow 3k=15\Rightarrow k=5$.