The 17th term of an A.P is 13 times the second term. Prove that the twelve term is five times the third term
you want to show that a+11d = 5(a+2d)
5(a+2d) = 5a + 10d
so, a+11d = 5a + 10d
d = 4a
The 17th term is a + 16d = a + 64a = 65a
5(a+2d) = 5(a + 8a) = 5(13a) = 65a
let s equal the 2nd term , and d equal the difference between adjacent terms
the 17th term equals the second term plus 15 differences
... 13 s = s + 15 d ... 12 s = 15 d ... 4 s = 5 d
the 12th term equals the second term plus 10 differences
... s + 10 d = 9 s
the 3rd term equals the 2nd term plus one difference ... s + d
... 5 (s + d) = 5 s + 5 d = 5 s + 4 s = 9 s
12th equals 5 times 3rd
"The 17th term of an A.P is 13 times the second term"
---> a+16d = 13(a+d)
a + 16d = 13a + 13d
12a = 3d
d =4a
Prove that the twelve term is five times the third term
LS = 12 th term = a + 11d = a + 44a = 45a
RS = 5(third term) = 5(a+2d)
= 5a + 10d
= 5a + 40a = 45a = LS
Done!
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To prove that the twelfth term is five times the third term in the arithmetic progression (A.P.), we need to find a relationship between the given terms.
Let's assume that the common difference is 'd' and the second term is 'a'.
So, the second term of the arithmetic progression (A.P.) can be expressed as: a + d.
Given that the 17th term is 13 times the second term, it can be written as: 13(a + d).
To find the twelfth term, we can use the formula for a term in an A.P.:
nth term = a + (n - 1)d, where n represents the position of the term.
Substituting n = 12 into this formula, we have:
12th term = a + (12 - 1)d
= a + 11d
Now, let's compare the 12th term (a + 11d) with the third term (a + 2d):
To prove that the twelfth term is five times the third term, we need to show that: a + 11d = 5(a + 2d).
Let's simplify this equation:
a + 11d = 5a + 10d
11d - 10d = 5a - a
d = 4a
Since 'd' is equal to 4a, we can substitute this value of 'd' into the equation for the 12th term:
12th term = a + 11d
= a + 11(4a)
= a + 44a
= 45a
We also need to calculate the third term:
3rd term = a + 2d
= a + 2(4a)
= a + 8a
= 9a
Now, by comparing the twelfth term (45a) with the third term (9a), we can see that the twelfth term is indeed five times the third term:
45a = 5(9a)
Hence, we have proved that the twelfth term is five times the third term in the given arithmetic progression.