Question

plz help
the question is if 9th term of an ap is 0 prove it's 29th term is double the 19th term.
plz tell is this the right way of solving

Answer 1

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Question :- If 9th term of an ap is 0 prove it's 29th term is double the 19th term.

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Step I

Write Given ,

$t_{9} = 0$

To prove :- $t_{29}$ = 2$t_{19}$

Step II

Use nth term formula for RHS

$t_{n}$ = a +( n - 1 ) d

$t_{9}$ = a + ( 9 - 1 ) d

0 = a + 8d ---- Equation (1)

Step III

Use the nth term formula on LHS

LHS = $t_{29}$

a + ( 29- 1 ) d

a + 28 d

a + 8 d+ 20 d

0 + 20d

$\boxed{A = 20d}$

Step IV

We have to prove RHS = 20

RHS = 2$t_{19}$

2 [ a +( 19 - 1) d]

2[ a + 18d ]

2 [ a + 8d + 10d ]

We know , the Value of a + 8d is 0

2 [ 0 + 10d ]

= $\boxed{20d}$

LHS = RHS

Hence , Proved

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@Brainlestuser

Answer 2

t^9 = 0
​

​

​

$Step II$

Use nth term formula for the RHS

t ^n = a +( n - 1 ) d

​
t^9 = a + ( 9 - 1 ) d

0 = a + 8d ---- Equation (1)

$Step III$

Use the nth term formula on LHS

LHS = t^29

​

a + ( 29- 1 ) d

a + 28 d

a + 8 d+ 20 d

0 + 20d

A = 20d

$Step IV$

to prove RHS = 20

RHS = 2t^19

​

2 [ a +( 19 - 1) d]

2[ a + 18d ]

2 [ a + 8d + 10d ]

We know , the Value of a + 8d is 0

2 [ 0 + 10d ]

= 20d

​
=>$L.H.S = R. H. S$

Hence Proved.....

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