if 9th term of an A.P is 0.prove that its 29th term is double the 19th term.
Calculate the 16th term in the geometric sequence: -1,3,-9,27
To prove that the 29th term of the arithmetic progression (A.P.) is double the 19th term when the 9th term is 0, we need to use the formula for the nth term of an A.P.
The formula to find the nth term of an A.P. is:
aₙ = a₁ + (n-1)d
where aₙ is the nth term, a₁ is the first term, n is the position of the term, and d is the common difference of the A.P.
Given that the 9th term (a₉) of the A.P. is 0, we can substitute these values into the formula:
0 = a₁ + (9-1)d
0 = a₁ + 8d
Now, let's calculate the 19th term and the 29th term using the same formula.
For the 19th term (a₁₉):
a₁₉ = a₁ + (19-1)d
= a₁ + 18d
For the 29th term (a₂₉):
a₂₉ = a₁ + (29-1)d
= a₁ + 28d
To prove that a₂₉ is double the value of a₁₉, we need to show that:
a₂₉ = 2 * a₁₉
Substituting the expressions for a₁₉ and a₂₉:
a₁ + 28d = 2(a₁ + 18d)
Expanding 2(a₁ + 18d):
a₁ + 28d = 2a₁ + 36d
Moving like terms to one side of the equation:
a₁ - 2a₁ = 36d - 28d
-a₁ = 8d
Dividing both sides of the equation by -1:
a₁ = -8d
Now, substituting the value of a₁ into the equation a₁ + 8d = 0 (from earlier):
-8d + 8d = 0
0 = 0
Since both sides are equal, the given A.P. satisfies the condition, and we have proved that the 29th term (a₂₉) is double the 19th term (a₁₉) when the 9th term is 0.