हिंदी

In an AP given a = 3, n = 8, Sn = 192, find d. - Mathematics

Advertisements
Advertisements

प्रश्न

 In an AP given a = 3, n = 8, Sn = 192, find d.

Let there be an A.P. with the first term 'a', common difference 'd'. If an a denotes in nth term and Sn the sum of first n terms, find.

d, if a = 3, n = 8 and Sn = 192

योग
Advertisements

उत्तर १

Given that, a = 3, n = 8, Sn = 192

`S_n = n/2 [2a+(n-1)d]`

`192 = 8/2[2xx3+(8-1)d]`

192 = 4 [6 + 7d]

48 = 6 + 7d

42 = 7d

`d = 42/7`

d = 6

shaalaa.com

उत्तर २

Here, we have an A.P. whose first term (a), the sum of first n terms (Sn) and the number of terms (n) are given. We need to find the common difference (d).

Here,

First term (a) = 3

Sum of n terms (Sn) = 192

Number of terms (n) = 8

So here we will find the value of n using the formula, an = a + (a - 1)d

So, to find the common difference of this A.P., we use the following formula for the sum of n terms of an A.P

`S_n = n/2 [2a + (n -1)d]`

Where; a = first term for the given A.P.

d = common difference of the given A.P.

n = number of terms

So, using the formula for n = 8, we get,

`S_8 = 8/2 [2(3) + (8 - 1)(d)]`

192 = 4[6 + (7) (d)]

192 = 24 + 28d

28d = 192 - 24

Further solving for d

`d = 168/28`

d = 6

Therefore, the common difference of the given A.P. is d = 6.

shaalaa.com
  क्या इस प्रश्न या उत्तर में कोई त्रुटि है?
अध्याय 5: Arithmetic Progressions - Exercise 5.3 [पृष्ठ ११२]

APPEARS IN

एनसीईआरटी Mathematics [English] Class 10
अध्याय 5 Arithmetic Progressions
Exercise 5.3 | Q 3.09 | पृष्ठ ११२
एमएल अग्रवाल Understanding Mathematics [English] Class 10 ICSE
अध्याय 9 Arithmetic and Geometric Progressions
Exercise 9.3 | Q 4.5
आरडी शर्मा Mathematics [English] Class 10
अध्याय 5 Arithmetic Progression
Exercise 5.6 | Q 56. 3

संबंधित प्रश्न

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


Find the sum given below:

`7 + 10 1/2 + 14 + ... + 84`


Which term of the progression 20, 19`1/4`,18`1/2`,17`3/4`, ... is the first negative term?


Show that the sum of all odd integers between 1 and 1000 which are divisible by 3 is 83667.


Find the sum 2 + 4 + 6 ... + 200


How many terms of the A.P. : 24, 21, 18, ................ must be taken so that their sum is 78?


If the 8th term of an A.P. is 37 and the 15th term is 15 more than the 12th term, find the A.P. Also, find the sum of first 20 terms of A.P.


If the sum of first n terms is  (3n+  5n), find its common difference.


The next term of the A.P. \[\sqrt{7}, \sqrt{28}, \sqrt{63}\] is ______.


Draw a triangle PQR in which QR = 6 cm, PQ = 5 cm and times the corresponding sides of ΔPQR?


Find the first term and common difference for  the A.P.

0.6, 0.9, 1.2,1.5,...


Find the sum of all even numbers between 1 and 350.

Kargil’s temperature was recorded in a week from Monday to Saturday. All readings were in A.P. The sum of temperatures of Monday and Saturday was 5°C more than sum of temperatures of Tuesday and Saturday. If temperature of Wednesday was –30° celsius then find the temperature on the other five days.


Choose the correct alternative answer for  the following question .

First four terms of an A.P. are ....., whose first term is –2 and common difference is –2.


In an A.P. the first term is – 5 and the last term is 45. If the sum of all numbers in the A.P. is 120, then how many terms are there? What is the common difference?


If the sum of first p terms of an A.P. is equal to the sum of first q terms then show that the sum of its first (p + q) terms is zero. (p ≠ q)


Find the sum:  1 + 3 + 5 + 7 + ... + 199 .


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


If 18th and 11th term of an A.P. are in the ratio 3 : 2, then its 21st and 5th terms are in the ratio


An article can be bought by paying Rs. 28,000 at once or by making 12 monthly installments. If the first installment paid is Rs. 3,000 and every other installment is Rs. 100 less than the previous one, find:

  1. amount of installments paid in the 9th month.
  2. total amount paid in the installment scheme.

Show that a1, a2, a3, … form an A.P. where an is defined as an = 3 + 4n. Also find the sum of first 15 terms.


In an A.P. sum of three consecutive terms is 27 and their products is 504. Find the terms. (Assume that three consecutive terms in an A.P. are a – d, a, a + d.)


Write the formula of the sum of first n terms for an A.P.


Determine the sum of first 100 terms of given A.P. 12, 14, 16, 18, 20, ......

Activity :- Here, a = 12, d = `square`, n = 100, S100 = ?

Sn = `"n"/2 [square + ("n" - 1)"d"]`

S100 = `square/2 [24 + (100 - 1)"d"]`

= `50(24 + square)`

= `square`

= `square`


In an AP, if Sn = n(4n + 1), find the AP.


If Sn denotes the sum of first n terms of an AP, prove that S12 = 3(S8 – S4)


Find the sum of all odd numbers between 351 and 373.


Find the sum of first 16 terms of the A.P. whose nth term is given by an = 5n – 3.


The 5th term and the 9th term of an Arithmetic Progression are 4 and – 12 respectively.

Find:

  1. the first term
  2. common difference
  3. sum of 16 terms of the AP.

The nth term of an A.P. is 6n + 4. The sum of its first 2 terms is ______.


Share
Notifications

Englishहिंदीमराठी


      Forgot password?
Use app×