English

If the sum of three consecutive terms of an increasing A.P. is 51 and the product of the first and third of these terms is 273, then the third term is

Advertisements
Advertisements

Question

If the sum of three consecutive terms of an increasing A.P. is 51 and the product of the first and third of these terms is 273, then the third term is

Options

  •  13

  • 9

  •  21

  • 17

MCQ
Advertisements

Solution

In the given problem, the sum of three consecutive terms of an A.P is 51 and the product of the first and the third terms is 273.

We need to find the third term.

Here,

Let the three terms be  (a-d),a,(a + d) where, a is the first term and d is the common difference of the A.P

So,

(a - d) + a + ( a + d) = 51 

                            3a = 51

                             `a = 51/3`

                              a = 17 

Also,

( a - d) ( a + d ) = 273 

             a2 - d2 = 273                [ Using  a2 - d2 = (a + b ) (a - b)]

             172 - d2 = 273

             289 - d2 = 273

Further solving for d,

 

Now, it is given that this is an increasing A.P. so cannot be negative.

So, d = 4

Substituting the values of a and d in the expression for the third term, we get,

Third term = a + d

So,

a + d = 17 + 4 

          = 21 

Therefore, the third term is  a3 = 21 

shaalaa.com
  Is there an error in this question or solution?
Chapter 5: Arithmetic Progressions - Exercise 5.8 [Page 57]

APPEARS IN

R.D. Sharma Mathematics [English] Class 10
Chapter 5 Arithmetic Progressions
Exercise 5.8 | Q 7 | Page 57

RELATED QUESTIONS

Ramkali required Rs 2,500 after 12 weeks to send her daughter to school. She saved Rs 100 in the first week and increased her weekly saving by Rs 20 every week. Find whether she will be able to send her daughter to school after 12 weeks.

What value is generated in the above situation?


The sum of the first p, q, r terms of an A.P. are a, b, c respectively. Show that `\frac { a }{ p } (q – r) + \frac { b }{ q } (r – p) + \frac { c }{ r } (p – q) = 0`


If the sum of first m terms of an A.P. is the same as the sum of its first n terms, show that the sum of its first (m + n) terms is zero


If the ratio of the sum of the first n terms of two A.Ps is (7n + 1) : (4n + 27), then find the ratio of their 9th terms.


Find the sum to n term of the A.P. 5, 2, −1, −4, −7, ...,


Find the sum 2 + 4 + 6 ... + 200


If the numbers a, 9, b, 25 from an AP, find a and b.


Write the next term of the AP `sqrt(2) , sqrt(8) , sqrt(18),.........`

 


Write an A.P. whose first term is a and common difference is d in the following.

a = –3, d = 0


Simplify `sqrt(50)`


If the sum of first n terms of an A.P. is  \[\frac{1}{2}\] (3n2 + 7n), then find its nth term. Hence write its 20th term.

 
 

Write the nth term of the \[A . P . \frac{1}{m}, \frac{1 + m}{m}, \frac{1 + 2m}{m}, . . . .\]

 

If the sum of n terms of an A.P. is 2n2 + 5n, then its nth term is


Q.1


 Q.10


Q.12


Q.19


If the sum of the first m terms of an AP is n and the sum of its n terms is m, then the sum of its (m + n) terms will be ______.


Which term of the AP: –2, –7, –12,... will be –77? Find the sum of this AP upto the term –77.


The sum of A.P. 4, 7, 10, 13, ........ upto 20 terms is ______.


Share
Notifications

Englishहिंदीमराठी


      Forgot password?
Use app×