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An Arithmetic progression has 13 terms whose sum is 143. The third term is 5 so the first term si
- a)4
- b)7
- c)9
- d)2
Correct answer is option 'D'. Can you explain this answer?
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An Arithmetic progression has 13 terms whose sum is 143. The third ter...
Given Information:
- Arithmetic progression has 13 terms
- Sum of those 13 terms is 143
- Third term of the AP is 5
To Find:
- First term of the AP
Solution:
Let the first term of the AP be 'a' and the common difference between the terms be 'd'.
We know that the sum of an AP is given by the formula:
Sum = (number of terms / 2) * (first term + last term)
Using this formula, we can find the last term of the AP as follows:
143 = (13 / 2) * (a + last term)
143 = 6.5a + 6.5(last term)
143 = 6.5(a + last term) --- (1)
Also, we know that the third term of the AP is given by:
3rd term = a + 2d
Substituting the given value of the third term (which is 5), we get:
5 = a + 2d --- (2)
Solving equations (1) and (2), we get:
last term = a + 12d = 11a - 55
Substituting this value of last term in equation (1), we get:
143 = 6.5(a + 11a - 55)
143 = 6.5(12a - 55)
22 = 12a - 55
12a = 77
a = 6.4167 (approx.)
Since the first term of the AP has to be a whole number, we can round off the value of 'a' to the nearest integer, which is 6.
Therefore, the first term of the AP is 6.
Answer: Option (d) 2.
- Arithmetic progression has 13 terms
- Sum of those 13 terms is 143
- Third term of the AP is 5
To Find:
- First term of the AP
Solution:
Let the first term of the AP be 'a' and the common difference between the terms be 'd'.
We know that the sum of an AP is given by the formula:
Sum = (number of terms / 2) * (first term + last term)
Using this formula, we can find the last term of the AP as follows:
143 = (13 / 2) * (a + last term)
143 = 6.5a + 6.5(last term)
143 = 6.5(a + last term) --- (1)
Also, we know that the third term of the AP is given by:
3rd term = a + 2d
Substituting the given value of the third term (which is 5), we get:
5 = a + 2d --- (2)
Solving equations (1) and (2), we get:
last term = a + 12d = 11a - 55
Substituting this value of last term in equation (1), we get:
143 = 6.5(a + 11a - 55)
143 = 6.5(12a - 55)
22 = 12a - 55
12a = 77
a = 6.4167 (approx.)
Since the first term of the AP has to be a whole number, we can round off the value of 'a' to the nearest integer, which is 6.
Therefore, the first term of the AP is 6.
Answer: Option (d) 2.
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Community Answer
An Arithmetic progression has 13 terms whose sum is 143. The third ter...
Okay so
S(13)=143
S(13)= (n/2)(2a+(n-1)d)
=(13/2)*(2a-12d)
=13*(a+6d)
=13a+78d=143
divide both sides by 13
a+6d=11………..(1)
T(3) = a+2d=5……..(2)
subtract (2) from (1)
4d=6
d=3/2
substitute d in any of the equations…(am using 2)
a+2(3/2)=5
a+3=5
a=2
S(13)=143
S(13)= (n/2)(2a+(n-1)d)
=(13/2)*(2a-12d)
=13*(a+6d)
=13a+78d=143
divide both sides by 13
a+6d=11………..(1)
T(3) = a+2d=5……..(2)
subtract (2) from (1)
4d=6
d=3/2
substitute d in any of the equations…(am using 2)
a+2(3/2)=5
a+3=5
a=2
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An Arithmetic progression has 13 terms whose sum is 143. The third term is 5 so the first term sia)4b)7c)9d)2Correct answer is option 'D'. Can you explain this answer?
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