Question:
If the coefficients of \(T_r, T_{r+1}, T_{r+2}\) terms of \((1+x)^{14}\) are in A.P., then \(r =\)
यदि \((1+x)^{14}\) के पद \(T_r, T_{r+1}, T_{r+2}\) के गुणांक A.P. में हैं, तो \(r =\)
Options:
- 6
- 7
- 8
- 9
- 6
- 7
- 8
- 9
Correct Answer: d)
Explanation:
The coefficients of \(T_r, T_{r+1}, T_{r+2}\) in \((1+x)^{14}\) are \({}^{14}C_{r-1}, {}^{14}C_r, {}^{14}C_{r+1}\).
Since they are in A.P.:
\[2 \cdot {}^{14}C_r = {}^{14}C_{r-1} + {}^{14}C_{r+1}\]
Dividing by \({}^{14}C_r\):
\[2 = \frac{r}{14 – r + 1} + \frac{14 – r}{r + 1} = \frac{r}{15 – r} + \frac{14 – r}{r + 1}\]
\[2(15 – r)(r + 1) = r(r + 1) + (14 – r)(15 – r)\]
\[2(15 + 14r – r^2) = (r^2 + r) + (210 – 29r + r^2)\]
\[30 + 28r – 2r^2 = 2r^2 – 28r + 210\]
\[4r^2 – 56r + 180 = 0 \implies r^2 – 14r + 45 = 0\]
\[(r – 5)(r – 9) = 0 \implies r = 5 \text{ or } r = 9\]
Among the options, \(r = 9\) is given.
Hence, the correct option is (d).
गुणांक \({}^{14}C_{r-1}, {}^{14}C_r, {}^{14}C_{r+1}\) समांतर श्रेणी में हैं:
\[2 \cdot {}^{14}C_r = {}^{14}C_{r-1} + {}^{14}C_{r+1}\]
हल करने पर द्विघात समीकरण प्राप्त होता है:
\[r^2 – 14r + 45 = 0 \implies (r – 5)(r – 9) = 0\]
अतः \(r = 5\) या \(r = 9\)। विकल्पों में 9 दिया गया है।
अतः सही विकल्प (d) है।