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In an A.P. if...

Question

In an $A . P .$ if $a_{2} + a_{5} - a_{3} = 10$ and $a_{2} + a_{9} = 17,$ then the value of $a + d$ is

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Solution

$a_{2} + a_{5} - a_{3} = 10 \Rightarrow (a + d) + (a + 4 d) - (a + 2 d) = 10 \Rightarrow a + 3 d = 10 \dots (i)$

$a_{2} + a_{9} = 17 \Rightarrow (a + d) + (a + 8 d) = 17 \Rightarrow 2 a + 9 d = 17 \dots (i i)$
Solving both the equations, we get
$a = 13, d = - 1 ∴ a + d = 12$

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a_{1} + a_{5} + a_{9} + . . . + a_{29} = 64

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a

d

a_{2} + a_{5} - a_{3} = 10

a_{2} + a_{9} = 17

A . P .

a_{1}, a_{2}, a_{3}, . . . . . ., a_{n}, . . . . .

a_{1} + a_{3} + a_{5} = - 12

a_{1} a_{2} a_{3} = 8

a_{2} + a_{4} + a_{6}

a_{1}, a_{2}, a_{3}, . . . . . . . a_{n}

a_{1} = 0

d \neq 0

\frac{a_{3} - a_{2}}{a_{2}} + \frac{a_{4} - a_{2}}{a_{3}} + \frac{a_{5} - a_{2}}{a_{4}} . . . . . . . . . . . . \frac{a_{n} - a_{2}}{a_{n - 1}}

a_{1}, a_{2}, a_{3} . . . . . . . . . a_{n},

a_{1} + a_{3} + a_{5} = - 12

a_{1} a_{2} a_{3} = 8

a_{2} + a_{4} + a_{6}

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