You visited us 1 times! Enjoying our articles? Unlock Full Access!

Area of Square

In an A.P i...

Question

In an $A . P$ if $a = 2$ and $t_{5} = 12$ , find $t_{2}, t_{3}, t_{4}$ .

Open in App

Solution

$a = 2 t_{5} = 12 \Rightarrow a + 4 d = 12 \Rightarrow 2 + 4 d = 12 \Rightarrow 4 d = 12 - 2 \Rightarrow 4 d = 10 \Rightarrow d = \frac{10}{4} ∴ d = \frac{5}{2} t_{2} = a + d = 2 + \frac{5}{2} = \frac{4 + 5}{2} = \frac{9}{2} = 4.5 t_{3} = a + 2 d = 2 + 2 \times \frac{5}{2} = 2 + 5 = 7 t_{4} = a + 3 d = 2 + 3 \times \frac{5}{2} = 2 + \frac{15}{2} = \frac{4 + 15}{2} = \frac{19}{2} = 9.5 t_{2} = 4.5 o r 4 \frac{1}{2} t_{3} = 7 t_{4} = 9.5 o r 9 \frac{1}{2}$

Suggest Corrections

Similar questions

\begin{matrix} T_{5} = 81 T_{2} = 3 \end{matrix}

T_{3}

T_{4}

t_{1}, t_{2}, t_{3}, t_{4}, t_{5}

D =

∣ ∣ ∣ ∣ \begin{matrix} t_{2} t_{3} & t_{2} & t_{1} t_{3} t_{4} & t_{3} & t_{2} t_{4} t_{5} & t_{4} & t_{3} \end{matrix} ∣ ∣ ∣ ∣

2 d^{4}

S_{1} = T_{1} + T_{2} + T_{3} + \dots + T_{n}

S_{2} = T_{1} + T_{3} + T_{5} + \dots + T_{n}, t h e n S_{1} / S_{2} =

T_{0}, T_{1}, T_{2}, . . . . . T_{n}

(x + a)^{n}

(T_{0} - T_{2} + T_{4} - T_{6} + . . . . .)^{2} + (T_{1} - T_{3} + T_{5} + . . . . .)^{2}

S_{1} = T_{1} + T_{2} + T_{3} + . . . + T_{n}

[n

]

S_{2} = T_{1} + T_{3} + T_{5} + . . . + T_{n}

\frac{S_{1}}{S_{2}} =

Join BYJU'S Learning Program