Question

if the 9 th term of an A.P. is 0. Then proove that its 29 th term is double than the 19 th term.​

Answer 1

☆ ANSWER:-

Given:-

• 9th term of an AP is 0.

To Prove:-

• 29th term is double than 19th term.

Formula Used:-

$\tt\leadsto a_n = a+(n-1)d.$

Where,

• n = number of terms.

• $\tt a_n$ = nth term

• a = First Term.

• d = Common Difference.

Now, ATQ:-

$\\ \sf \mapsto a_{9} = 0 \\ \\ \sf \mapsto a+(9-1)d = 0. \\ \\ \sf \mapsto a+8d = 0. \\ \\ \sf \mapsto a=-8d ...(1).$

So,

$\\ \sf \mapsto a_{19} = a+(19-1)d. \\ \\ \sf \mapsto a_{19} = a+18d. \\ \\ \sf \mapsto a_{19} = -8d+18d \\ \\ \sf [From\:\:(1)] \\ \\ \sf\mapsto a_{19} = 10d.$

Similarly,

$\\ \sf \mapsto a_{29} = a+(29-1)d. \\ \\ \sf \mapsto a_{29} = a+28d. \\ \\ \sf \mapsto a_{29} = -8d+28d. \\ \\ \sf [From \:\:(1)] \\ \\ \sf\mapsto a_{29} = 20d.$

So clearly we can see that,

$\sf \implies a_{19} = 10d. \\ \\ \sf \implies 2\times a_{19} = 2\times 10d. \\ \\ \sf\implies 2\times a_{19} = 20d = a_{29}. \\ \\ \sf\implies a_{29} = 2\times a_{19}.$

Hence Proved.