The 4th term of an A.P. is zero. Prove that the 25th term of the A.P. is three times its 11th term.
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The 4th term of an A.P. is zero. Prove that the 25th term of the A.P. is three times its 11th term.
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Verified answer
Step-by-step explanation:
Given :-
To prove :-
Solution :-
Formula used :
[tex]{\boxed{\sf{T_n=a+(n-1)d}}}[/tex]
Now find the 4th term of the A.P.
★[tex]\sf{T_4=a+(4-1)d}[/tex]
[tex]\to\sf{T_4=a+3d}[/tex]
According to the question,
a + 3d = 0
→ a = -3d............(i)
Now find the 25th term and 11th term of the A.P.
★ [tex]\sf{T_{25}=a+(25-1)d}[/tex]
[tex]\to\sf{T_{25}=a+24d}[/tex]
And,
★ [tex]\sf{T_{11}=a+(11-1)d}[/tex]
[tex]\to\sf{T_{11}=a+10d}[/tex]
Then,
[tex]{\sf{25th\:term=3(11th\:term)}}[/tex]
→ a+24d = 3(a+10d)
→ -3d + 24d = 3(-3d+10d)
→ 21d = 3 × 7d
→ 21d = 21d
Hence proved that the 25th term of the A.P. is 3 times its 11th term.