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Class 12 Tuition > Learn Class XI-XII Tuition (PUC) > The sum of first four terms of an A.P. is 56 . The...
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The sum of first four terms of an A.P. is 56 . The sum of last four terms is 112. If its first term is 11 then find the number of terms ?

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Let the first term be 'a' Given:- a=11. Also, it is given that sum of its 1st four terms is 56, which means a + a + d + a + 2d + a + 3d=56 => 4a+6d = 56=> 44+6d = 56 Therefore, d=2 Again given that sum of last four terms is 112. a + (n-1) d + a + (n-2)d +...
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Let the first term be 'a' Given:- a=11. Also, it is given that sum of its 1st four terms is 56, which means a + a + d + a + 2d + a + 3d=56 => 4a+6d = 56=> 44+6d = 56 Therefore, d=2 Again given that sum of last four terms is 112. a + (n-1) d + a + (n-2)d + a + (n-3)d + a + (n-4)d = 112 putting a=11 and d=2 => 44+8n-20 = 112 So, solving this we get n=11. There are 11 terms in this A.P The series is 11,13,15,17,19,21,23,25,27,29,31. 11+13+15+17=56 31+29+27+25=112 read less
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The proof to the Rational Numbers question posted, is as follows:- A number 'n' can be partitioned without repetition in 'n-1' ways. So total number of partitions till 'n-1' let us call this j = 1/2 * (n-2) (n-1)...since there are n-2 partitions for 'n-1' Let this 'n'= p+q As the 1st term of 'n'...
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The proof to the Rational Numbers question posted, is as follows:- A number 'n' can be partitioned without repetition in 'n-1' ways. So total number of partitions till 'n-1' let us call this j = 1/2 * (n-2) (n-1)...since there are n-2 partitions for 'n-1' Let this 'n'= p+q As the 1st term of 'n' stars with q=1 p/q will be represented by (j+q)th term. read less
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The proof to the Rational Numbers question posted, is as follows:- A number 'n' can be partitioned without repetition in 'n-1' ways. So total number of partitions till 'n-1' let us call this j = 1/2 * (n-2) (n-1)...since there are n-2 partitions for 'n-1' Let this 'n'= p+q As the 1st term of 'n'...
read more
The proof to the Rational Numbers question posted, is as follows:- A number 'n' can be partitioned without repetition in 'n-1' ways. So total number of partitions till 'n-1' let us call this j = 1/2 * (n-2) (n-1)...since there are n-2 partitions for 'n-1' Let this 'n'= p+q As the 1st term of 'n' starts with q=1 p/q will be represented by (j+q)th term. read less
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Let the AP be a + (a +d)+ (a+2d)+ (a +3d) +....... .................... + a +(n-1)d + a +(n -2)d + a +(n-3)d +a +(n-4)d a + (a +d)+ (a+2d)+ (a +3d) = 56 4d + 6d = 56 since a = 11, d = 2 sum of the last 4 terms = 112 (n-1)d + a +(n -2)d + a +(n-3)d +a +(n-4)d =112 4a + (4n-10)d = 112 plug...
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Let the AP be a + (a +d)+ (a+2d)+ (a +3d) +....... .................... + a +(n-1)d + a +(n -2)d + a +(n-3)d +a +(n-4)d a + (a +d)+ (a+2d)+ (a +3d) = 56 4d + 6d = 56 since a = 11, d = 2 sum of the last 4 terms = 112 (n-1)d + a +(n -2)d + a +(n-3)d +a +(n-4)d =112 4a + (4n-10)d = 112 plug in a = 11 and d= 2 that gives us n = 11 terms read less
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Hardworking

11 terms
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Math Educator for Std.11th ,12th , Engineering Entrance and Degree Level with 11+ Years Experience

number of terms = 11
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Math analyst

11 + 11+d +11+2d +11+3d = 44 +6d = 56. 6d = 56-44 = 12; d = 2. 11+(n-1)d +11 + (n-2)d + 11 +(n-3)d + 11 + (n-4)d = 112; 44 + (4n - 10)*2 = 112; (4n-10)=(112-44)/2 = 68/2 = 34 4n = 44; n =11. wow!
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Spoken English, BTech Tuition, C Language, C++ Language, CAD etc., with 4 years of experience

11
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Maths Teacher

First term is 11.Sum of first 4 terms is 4a+6d=56. That is 44+6d=56.So d=2 Sum of last four terms is 4a+d=4a+(4n-10)d=112 (4n-10)2= 112-44=68. 4n-10=34. 4n=44 Hence n=11
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Let a = first term, d = common difference and n = number of terms Then, a + a + d + a + 2d + a + 3d = 56 4a + 6d = 56 44 + 6d = 56 (since a = 11) d = 2 Now, 4a + d {(n-1) + (n-2) + (n-3) + (n-4)} = 112 44 + 2 (4n - 10) = 112 4n - 10 = 34 4n = 44 n = 11 Hence, there are 11 terms.
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