The $4^{th}$ term of an A.P. is zero. Prove that the $25^{th}$ term of the A.P. is three times its $11^{th}$ term.
Given: The 4th term of an A.P. is zero.
To do: To Prove that the 25th term of the A.P. is three times its 11th term.
Solution:
4th term of an $A.P.\ =\ a_{4} =0$
$\therefore \ a\ +\ ( 4\ –\ 1) d\ =0$
$\therefore \ a\ +\ 3d\ =\ 0$
$\therefore \ a\ =\ –3d\ \ \ \ \ \ \dotsc .( 1)$
25th term of an A.P. , $a_{25}$
$=a+( 25\ –\ 1) d$
$=–3d\ +\ 24d\ \ \ \ \ \ \ \ \ \ \dotsc .[ From\ ( 1)]$
$=21d$
11th term of the given A.P. $a_{11} .=a+( n-1) d$
$=-3d+( 11-1) d$
$=-3d+10d$
$=7d$
3 times 11th term of an A.P. $=3a_{11} =3\times 7d=21d$
i.e., the $25^{th}$ term of the A.P. is three times its $11^{th}$ term.
Related Articles
- If $7$ times of the $7^{th}$ term of an A.P. is equal to $11$ times its $11^{th}$ term, then find its $18^{th}$ term.
- If $(m + 1)$th term of an A.P. is twice the $(n + 1)$th term, prove that $(3m + 1)$th term is twice the $(m + n + 1)$th term.
- If the $2^{nd}$ term of an A.P. is $13$ and the $5^{th}$ term is $25$, what is its $7^{th}$ term?
- If the \( 9^{\text {th }} \) term of an AP is zero, prove that its \( 29^{\text {th }} \) term is twice its \( 19^{\text {th }} \) term.
- The $14^{th}$ term of an A.P. is twice its $8^{th}$ term. If its $6^{th}$ term is $-8$, then find the sum of its first $20$ terms.
- In an AP five times of $5^{th}$ term is equal to ten times the $10^{th}$ term, show that its 15th term is zero.
- If $m^{th}$ term of an $A.P.$ is $\frac{1}{n}$ and $n^{th}$ term of another $A.P.$ is $\frac{1}{m}$. Then, show that $(mn)^{th}$ term is equal to $1$.
- The $9^{th}$ term of an AP is $499$ and its $499^{th}$ term is $9$. Which of its term is equal to zero.
- If $m$ times the $m^{th}$ term of an AP is equal to $n$ times its $n^{th}$ term. find the $( m+n)^{th}$ term of the AP.
- For an AP, if \( m \) times the mth term equals \( \mathrm{n} \) times the \( n \) th term, prove that \( (m+n) \) th term of the AP is zero. \( (m ≠ n) \).
- Prove that the $n^{th}$ term of an A.P. can't be $n^2+1$.
- If 9th term of an A.P. is zero, prove that its 29th term is double the 19th term.
- The 24th term of an A.P. is twice its 10th term. Show that its 72nd term is 4 times its 15th term.
- If an A.P. consists of $n$ terms with first term $a$ and $n$th term $l$ show that the sum of the $m$th term from the beginning and the $m$th term from the end is $(a + l)$
- Find the \( 20^{\text {th }} \) term of the AP whose \( 7^{\text {th }} \) term is 24 less than the \( 11^{\text {th }} \) term, first term being 12.
Kickstart Your Career
Get certified by completing the course
Get Started