if 3+k,15-k,5k+1 are in a.p. then find k.
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Answer:
Required numeric value of k is 13 / 4.
Step-by-step explanation:
Given terms( of an A.P. ) are 3 + k, 15 - k , 5k + 1.
Let 3 + k be nth term, 15 - k be ( n + 1 )th term and 5k + 1 e ( n + 2 )th term.
From the properties of arithmetic progressions: -
- nth term = a + ( n - 1 )d, where a is the is first term, n is the number of term and d is the common difference between the terms.
Then, let the first term of this AP be a and common difference between the term be d.
Now,
= > 3 + k = nth term
= > 3 + k = a + ( n - 1 )d ...( i )
= > 15 - k = ( n + 1 )th term
= > 15 - k = a + ( n + 1 - 1 )d
= > 15 - k = a + ( n )d ...( ii )
= > 5k + 1 = ( n + 2 ) th term
= > 5k + 1 = ( n + 2 - 1 )d
= > 5k + 1 = ( n + 1 )d ...( iii )
If we see, twice the ( n + 1 )th term is the sum of other terms, it simply means that the sum of a + ( n - 1 )d and a + ( n + 1 ) is a + ( n )d.
Thus,
= > [ a + ( n - 1 ) d ] + [ a + ( n + 1 )d ] = 2 ( a + nd )
Substituting the values from ( i )&( ii )&( iii ) : -
= > [ 3 + k ] + [ 5k + 1 ] = 2( 15 - k )
= > 3 + k + 5k + 1 = 30 - 2k
= > 6k + 4 = 30 - 2k
= > 6k + 2k = 30 - 4
= > 8k = 26
= > k = 26 / 8 = 13 / 4
Hence the required numeric value of k is 13 / 4.
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Solution : From the properties of arithmetic progressions, we know, in 3 consecutive terms, 2 x middle term = sum of remaining terms.
Therefore,
= > 3 + k + 5k + 1 = 2( 15 - k )
= > 3 + k + 5k + 1 = 30 - 2k
= > 6k + 4 = 30 - 2k
= > 6k + 2k = 30 - 4
= > 8k = 26
= > k = 26 / 8 = 13 / 4
Hence the required numeric value of k is 13 / 4.