If in a A-P- the third term is 3- the sixth term is 5 and nth term is 2033- then the value of n is

Given : T_3 = 3, T_6= 5, T_n = 2033 a+2d= 3 a+5d = 5 ⇒ d = 23⇒ a = 5 3 T_n = 2033 ⇒ a+(n 1)× d= 2033 ⇒ 5 3+(n 1)×23 = 2033 ⇒ 5+(n 1)× 2 = 203 ∴ n = 100 ...

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Standard XII

Mathematics

Series

If in a A.P. ...

Question

If in a A.P. the third term is $3$ , the sixth term is $5$ and $n^{t h}$ term is $\frac{203}{3}$ , then the value of $n$ is

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Solution

Given : $T_{3} = 3, T_{6} = 5, T_{n} = \frac{203}{3}$
$a + 2 d = 3 a + 5 d = 5 \Rightarrow d = \frac{2}{3} \Rightarrow a = \frac{5}{3} T_{n} = \frac{203}{3} \Rightarrow a + (n - 1) \times d = \frac{203}{3} \Rightarrow \frac{5}{3} + (n - 1) \times \frac{2}{3} = \frac{203}{3} \Rightarrow 5 + (n - 1) \times 2 = 203 ∴ n = 100$

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If in a A.P. the third term is 3, the sixth term is 5 and nth term is 2033, then the value of n is

Given : T3=3, T6=5, Tn=2033 a+2d=3a+5d=5⇒d=23⇒a=53Tn=2033⇒a+(n−1)×d=2033⇒53+(n−1)×23=2033⇒5+(n−1)×2=203∴n=100

If in a A.P. the third term is $3$ , the sixth term is $5$ and $n^{t h}$ term is $\frac{203}{3}$ , then the value of $n$ is