हिंदी

Find the 19th term of the following A.P.: 7, 13, 19, 25, ... - Algebra Mathematics 1

Advertisements
Advertisements

प्रश्न

Find the 19th term of the following A.P.:

7, 13, 19, 25, ...

योग
Advertisements

उत्तर

The given sequence is 7, 13, 19, 25, ....

Here,

First term = a = t1 = 7, t2 = 13, t3 = 19, t4 = 25, ....

Common difference = d = t2 − t1

= 13 – 7

= 6

To find the 19th term, we have to use the formula, i.e.,

tn = a + (n − 1)d

∴ t19 = 7 + (19 − 1) × 6 ...(On substituting value)

= 7 + 18 × 6

= 7 + 108

= 115

Hence, the 19th term of the progression is 115.

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 4 | पृष्ठ ६६

संबंधित प्रश्न

If the 9th term of an A.P. is zero, then prove that 29th term is double of 19th term.


Find the sum of  the following arithmetic series:

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


Find the sum of  the following arithmetic series:

34 + 32 + 30 +...+10


Decide whether the following sequence is an A.P., if so find the 20th term of the progression:

–12, –5, 2, 9, 16, 23, 30, ..............


Given Arithmetic Progression 12, 16, 20, 24, . . . Find the 24th term of this progression.


Find the 27th term of the following A.P.
9, 4, –1, –6, –11,...


Six year before, the age of mother was equal to the square of her son's age. Three year hence, her age will be thrice the age of her son then. Find the present ages of the mother and son.


Select the correct alternative and write it. 

If a share is at premium, then - 


For a given A.P. a = 3.5, d = 0, then tn = _______.


Choose the correct alternative answer for the following sub-question

If the third term and fifth term of an A.P. are 13 and 25 respectively, find its 7th term


Find tn if a = 20 आणि d = 3


Decide whether the given sequence 24, 17, 10, 3, ...... is an A.P.? If yes find its common term (tn)


How many two-digit numbers are divisible by 5?

Activity :-  Two-digit numbers divisible by 5 are, 10, 15, 20, ......, 95.

Here, d = 5, therefore this sequence is an A.P.

Here, a = 10, d = 5, tn = 95, n = ?

tn = a + (n − 1) `square`

`square` = 10 + (n – 1) × 5

`square` = (n – 1) × 5

`square` = (n – 1)

Therefore n = `square`

There are `square` two-digit numbers divisible by 5


Decide whether 301 is term of given sequence 5, 11, 17, 23, .....

Activity :-  Here, d = `square`, therefore this sequence is an A.P.

a = 5, d = `square`

Let nth term of this A.P. be 301

tn = a + (n – 1) `square`

301 = 5 + (n – 1) × `square`

301 = 6n – 1

n = `302/6` = `square`

But n is not positive integer.

Therefore, 301 is `square` the term of sequence 5, 11, 17, 23, ......


12, 16, 20, 24, ...... Find 25th term of this A.P.


Write the next two terms of the A.P.: `sqrt(27), sqrt(48), sqrt(75)`......


Share
Notifications

Englishहिंदीमराठी


      Forgot password?
Use app×