Question

sum of the first 15terms of an arithametic sequence is 735 5th term is 31 find the 7th term ?find the sum of nth term​

Answer 1

Answer:

Required 7th term is 43 and sum of n terms is 3n^2 + 4n.

Step-by-step-explanation:

From the properties of arithmetic progressions :

• nth term = a + ( n - 1 )d, where a is the first, n is the number of terms and d is the difference between the terms of an arithmetic progression.
• S${}_{n}=\dfrac{n}{2}$[ 2a + ( n - 1 )d ]

Let the first term of this progression be a and common difference between the terms be d.

It is given, 5th term of the AP is 31.

= > a + ( 5 - 1 )d = 31

= > a + 4d = 31

= > a = 31 - 4d ...( i )

Also, sum of first 15 terms is 735.

= > S${}_{15}=\dfrac{n}{2}$ x [ 2a + ( 15 - 1 )d ] = 735

= > $\dfrac{15}{2}$ x [ 2a + 14d ] = 735

= > $\dfrac{15}{2}$ x 2[ a + 7d ] = 735

= > 15[ a + 7d ] = 735

= > a + 7d = 735 / 15

= > a + 7d = 49

From ( i ), substituting the algebraic value of a :

= > 31 - 4d + 7d = 49

= > 31 + 3d = 49

= > 3d = 49 - 31

= > 3d = 18

= > d = 6

Substituting the numeric value of d in ( i ) :

= > a = 31 - 4d

= > a = 31 - 4( 6 )

= > a = 31 - 24

= > a = 7

Thus,

= > 7th term of this AP

= > a + ( 7 - 1 )d

= > a + 6d

= > 7 + 6( 6 )

= > 7 + 36

= > 43

= > Sum of n terms

= > $\dfrac{n}{2}$[ 2a + ( n - 1 )d ]

= > ( n / 2 ) [ 2( 7 ) + ( n - 1 )6 ]

= > ( n / 2 ) [ 14 + 6n - 6 ]

= > ( n / 2 ) [ 8 + 6n ]

= > ( n / 2 ) x 2[ 4 + 3n ]

= > n( 4 +3n )

= > 3n² + 4n

Hence the required 7th term is 43 and sum of n terms is 3n^2 + 4n.