First slide

Arithmetic progression

Question

$a_{1}, a_{2}, a_{3} \dots a_{n} are in A.P such that a_{n} = 100, a_{50} - a_{49} = 3 / 5 then 15^{th} term of A.P from the end is$

Moderate

A

$\frac{454}{5}$
B

$\frac{448}{5}$
C

$\frac{452}{5}$
D

$\frac{458}{5}$

Solution

$Let T_{15} is 15^{th} term from end$
$\begin{array}{l} T_{15} = a_{n} - 14 d = 100 - 14 (\frac{3}{5}) \\ = 100 - \frac{42}{5} = \frac{458}{5} \end{array}$

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