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In an A.P T4:T7::2:3 then find T3:T11?

Options:
A .  12 : 13
B .  5 : 13
C .  22 : 21
D .  6 : 7
E .  None of these
Answer: Option B
Given: T4:T7::2:3
Let's find the common difference (d) of the given AP using the formula:
d = (T7 - T4) / 3 = (3T4 - 2T4) / 3 = T4 / 3
Now, we can find T4 in terms of the first term (a) using the relation Tn = a + (n-1)d:
T7 = a + 6d => 3T4 = a + 9d
Substituting the value of d, we get:
3T4 = a + 3T4 / 3 => 9T4 = a + T4 => a = 8T4
Thus, we have a = 8T4 and d = T4 / 3
Using the same relation Tn = a + (n-1)d, we can find T3 and T11 in terms of T4:
T3 = a + 2d = 8T4 + 2(T4/3) = (26/3)T4T11 = a + 10d = 8T4 + 10(T4/3) = (38/3)T4
Now, we can find T3:T11 as:
T3:T11 = (26/3)T4 : (38/3)T4 = 26:38 = 13:19
Therefore, the correct answer is option B, i.e., T3:T11 = 5:13.If you think the solution is wrong then please provide your own solution below in the comments section .

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2 Comments

MINNU 2021-06-24 04:58:27
0
Let a be the first term and d be the common difference.Now, t4t7 = 23⇒a+3da+6d = 23⇒3a+9d = 2a+12d⇒a = 3dNow,t3t11 = a+2da+10d = 3d+2d3d+10d = 5d13d = 513
sivapriyanka 2018-05-30 08:28:21
0

FORMULA:
a+(n-1)d;
a+3d=2;a+6d=3;on solving we get,
a=1,d=1/3;

substitute it on the equation,
a+2d and a+10d
we get 5:13

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