T₂=1,T₁₂=-9 5.,Find AP. If Tn, Tm are as given below.
Answers
Dear Student,
Answer: AP is 10.6 ,1 , -8.6 , -18.2 , -27.8, -37.4 ...
Solution:
we know that nth element of AP is given as = a+ (n-1) d
T₂=1
i.e. 1=a+ ( 2-1) d
1 = a+d
a+d = 1 .....eq1
T₁₂=-95
i.e -95 = a+ (12-1)d
a+11d = -95 .......eq2
Subtract eq2 from eq1
a+d-a-11d = 1+95
-10 d = 96
d = -96/10
d = -9.6
put value of d in any of the equation
a+d = 1
a-9.6 = 1
a = 1+9.6
a = 10.6
First element is 10.6 , common difference is -9.6
AP is a, a+d ,a+2d, a+3d ...
10.6, 10.6-9.6, 10.6-9.6(2) , 10.6 - 9.6(3), 10.6 - 9.6(4),....
10.6 ,1 , -8.6 , -18.2 , -27.8, -37.4 ...
Hope it helps you.
We know that Tn = a + (n - 1) d
Given T2 = 1
= > 1 = a + (2 - 1) * d
= > 1 = a + d ----- (1)
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Given T12 = -95
= > -95 = a + (12 - 1) * d
= > -95 = a + 11d ----- (2)
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On solving (1) & (2), we get
a + d = 1
a + 11d = -95
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-10d = 96
d = 96/10
d = -9.6
Substitute d = -1.05 in (1), we get
= > a + d = 1
= > a - 9.6 = 1
= > a = 1 + 9.6
= > a = 10.6
Therefore, the given series will be 10.6, 10.6 - 9.6, 10.6 - 9.6(2).......
= > 10.6, 1, -8.6.......
Hope this helps!