हिंदी

In an A.P. 17th term is 7 more than its 10th term. Find the common difference.

Advertisements
Advertisements

प्रश्न

In an A.P. 17th term is 7 more than its 10th term. Find the common difference.

योग
Advertisements

उत्तर

In an A.P., 17th term is 7 more than its 10th term.  ...(Given)

The formula for nth term of an A.P. is tn = a + (n − 1)d

The 10th term can be written as,

t10 = a + (10 − 1)d 

t10 = a + 9d

The 17th term can be written as,

t17 = a + (17 − 1)d 

t17 = a + 16d 

Now, according to the question,

t17 = 7 + t10

Substituting the value of t17 and t10 in the above equation,

∴ a + 16d = 7 + a + 9d

∴ 16d = 7 + 9d

∴ 16d − 9d = 7

∴ 7d = 7

∴ d = `7/7` = 1

Hence, the common difference of the given AP is 1.

shaalaa.com
  क्या इस प्रश्न या उत्तर में कोई त्रुटि है?
अध्याय 3: Arithmetic Progression - Practice Set 3.2 [पृष्ठ ६६]

APPEARS IN

बालभारती Algebra Mathematics 1 [English] Standard 10 Maharashtra State Board
अध्याय 3 Arithmetic Progression
Practice Set 3.2 | Q 10 | पृष्ठ ६६

संबंधित प्रश्न

The first and the last terms of an AP are 8 and 65 respectively. If the sum of all its terms is 730, find its common difference.


Find the sum of first 22 terms of an AP in which d = 7 and 22nd term is 149.


In a potato race, a bucket is placed at the starting point, which is 5 m from the first potato and other potatoes are placed 3 m apart in a straight line. There are ten potatoes in the line.

A competitor starts from the bucket, picks up the nearest potato, runs back with it, drops it in the bucket, runs back to pick up the next potato, runs to the bucket to drop it in, and she continues in the same way until all the potatoes are in the bucket. What is the total distance the competitor has to run?

[Hint: to pick up the first potato and the second potato, the total distance (in metres) run by a competitor is 2 × 5 + 2 ×(5 + 3)]


Find the sum of the first 25 terms of an A.P. whose nth term is given by an = 2 − 3n.


Is 184 a term of the AP 3, 7, 11, 15, ….?


How many two-digits numbers are divisible by 3?

 


If a denotes the nth term of the AP 2, 7, 12, 17, … find the value of (a30 - a20 ).


Find four consecutive terms in an A.P. whose sum is 12 and sum of 3rd and 4th term is 14.

(Assume the four consecutive terms in A.P. are a – d, a, a + d, a +2d) 


Two A.P.’s are given 9, 7, 5, ... and 24, 21, 18, ... If nth term of both the progressions are equal then find the value of n and nth term.


Divide 207 in three parts, such that all parts are in A.P. and product of two smaller parts will be 4623.


Rs 1000 is invested at 10 percent simple interest. Check at the end of every year if the total interest amount is in A.P. If this is an A.P. then find interest amount after 20 years. For this complete the following activity.


The first term of an A. P. is 5 and the common difference is 4. Complete the following activity and find the sum of the first 12 terms of the A. P.

a = 5, d = 4, s12 = ?
`s_n = n/2 [ square ]`
`s_12 = 12/2 [10 +square]`
         `= 6 × square  `
         ` =square`


Find the sum of all 2 - digit natural numbers divisible by 4.


​The first and the last terms of an A.P. are 7 and 49 respectively. If sum of all its terms is 420, find its common difference. 


If the sums of n terms of two arithmetic progressions are in the ratio \[\frac{3n + 5}{5n - 7}\] , then their nth 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.


Find the sum of natural numbers between 1 to 140, which are divisible by 4.

Activity: Natural numbers between 1 to 140 divisible by 4 are, 4, 8, 12, 16,......, 136

Here d = 4, therefore this sequence is an A.P.

a = 4, d = 4, tn = 136, Sn = ?

tn = a + (n – 1)d

`square` = 4 + (n – 1) × 4

`square` = (n – 1) × 4

n = `square`

Now,

Sn = `"n"/2["a" + "t"_"n"]`

Sn = 17 × `square`

Sn = `square`

Therefore, the sum of natural numbers between 1 to 140, which are divisible by 4 is `square`.


If the sum of three numbers in an A.P. is 9 and their product is 24, then numbers are ______.


In an A.P., if Sn = 3n2 + 5n and ak = 164, find the value of k.


Share
Notifications

Englishहिंदीमराठी


      Forgot password?
Use app×