माना a _{1}, a _{2}, ldots ldots a _{30} एक स | vedclass

Name: PDF Generator App | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

$S = \sum\limits_{i - 1}^{30} {{a_i}} \,\,\,\,,T = \sum\limits_{i - 1}^{15} {{a_{2i - 1}}} \,\,\,\,\,,{a_5} = 27,S - 2T = 75$

Let ${a_i} = a +

Vedclass · Accepted Answer

$52$

Vedclass · Answer

$S = \sum\limits_{i - 1}^{30} {{a_i}} \,\,\,\,,T = \sum\limits_{i - 1}^{15} {{a_{2i - 1}}} \,\,\,\,\,,{a_5} = 27,S - 2T = 75$

Let ${a_i} = a + \left( {i - 1} ight)D$

$S = {a_1} + {a_2} + {a_3} + ........... + {a_{30}}$

$T = {a_1} + {a_3} + {a_5} + ........... + {a_{29}}$

$	herefore 2T = 2{a_1} + 2{a_3} + 2{a_5} + ........... + 2{a_{29}}$

$S - 2T = \left( {{a_2} - {a_1}} ight) + \left( {{a_4} - {a_3}} ight) + \left( {{a_6} - {a_5}} ight) + ........ + \left( {{a_{30}} - {a_{29}}} ight) = 75$

$ = 15D$

But $S - 2T = 75 \Rightarrow 15D = 75 \Rightarrow D = 5$

Now ${a_5} = 27 \Rightarrow a + 4D = 27$

$	herefore a = 27 - 20 \Rightarrow a = 7$

${a_{10}} = a + 9D$

$ = 7 + 45 = 52$

माना $a _{1}, a _{2}, \ldots \ldots a _{30}$ एक समांतर श्रेणी है. $S =\sum_{i=1}^{30} a _{i}$ तथा $T = \sum\limits_{i = 1}^{15} {{a_{2i - 1}}} $ यदि $a _{5}=27$ तथा $S -2 T =75$, तो $a _{10}$ बराबर है

Similar Questions

यदि $x,y,z$ समान्तर श्रेणी में हों तथा ${\tan ^{ - 1}}x,{\tan ^{ - 1}}y$, ${\tan ^{ - 1}}z$ भी समान्तर श्रेणी में हों, तब

यदि A.P. $a _{1} a _{2}, a _{3}, \ldots$ के प्रथम 11 पदों का योगफल $0\left(a_{1} \neq 0\right)$ है और A.P., $a_{1}, a_{3}, a_{5}, \ldots, a_{23}$ का योगफल $ka _{1}$ है, तो $k$ बराबर है -

यदि $1,\,\,{\log _9}({3^{1 - x}} + 2),\,\,{\log _3}({4.3^x} - 1)$ समान्तर श्रेणी में हों, तो $x$ का मान होगा