Question

15. श्रेणी ज्ञात करो जिसका पहला पद 3 है और 19वाँ पद 129 है।
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Answer 1

दिए गए प्रश्न से सही उत्तर है :

अनुक्रम तब है जब पहला पद 3 है और 19वाँ पद 129 है:

3, 9.14, 15.28 ,21.42, 27.56, 33.7,39.84 ,45.98,52.12, 58.26, 64.4,

70.54, 76.68, 82.82 , 88.96 ,95.1, 101.24,107.38, 113.52,119.66,129.8

दिया गया:

$a_{1}$=3

$a_{n}$=129

n=19

Solution:

श्रृंखला बनाने के लिए हमें सामान्य अंतर खोजना होगा।

हमारे पास अनुक्रम में अंतिम पद खोजने का एक सूत्र है।

$a_{n}$ = $a_{1}$ + (n - 1) x d

= nth term

$a_{1}$ = पहला पद

n = पद संख्या

d =सामान्य अंतर

हम इस सूत्र का उपयोग करने से अंतर पा सकते हैं 

$a_{n}$ = $a_{1}$ + (n - 1) x d

129=3+(19-1)×d

129=3+18×d

129=21×d

129/21=d

d=129/21

d=6.14

अब क्रम ढूंढते हैं

पहले पद

$a_{1}$

दूसरा पद :

$a_{2}$=$a_{1}$ + d

=3+6.14

=9.14

तीसरी पद :

$a_{3}$=$a_{1}$+ 2d

=3+2×6.14

=15.28

चौथी  पद:

$a_{4}$=$a_{1}$ + 3d

=3+3×6.14

=21.42

पांचवां पद:

$a_{5}$=$a_{1}$ + 4d

=3+4×6.14

=27.56

छठा पद:

$a_{6}$=$a_{1}$+5d

=3+5×6.14

=33.7

इस प्रकार हम इस अनुक्रम के सभी 19 पद प्राप्त कर सकते हैं

तो क्रम है

3, 9.14, 15.28 ,21.42, 27.56, 33.7,39.84 ,45.98,52.12, 58.26, 64.4,

70.54, 76.68, 82.82 , 88.96 ,95.1, 101.24,107.38, 113.52,119.66,129.8

Answer 2

Answer:

Step-by-step explanation:

The correct answer from the given question is :

The sequence is when the first term is 3 and the 19th term is 129 is:

3, 9.14, 15.28,21.42, 27.56, 33.7,39.84, 45.98,52.12, 58.26, 64.4,

70.54, 76.68, 82.82, 88.96,95.1, 101.24,107.38, 113.52, 119.66,129.8

Given:

a1 = 3

aa = 129

n = 19

Solution:

To form the series we need to find the common difference Will happen.

We have a way of finding the last term in the sequence is the formula.

aa = a1+ (n - 1) x d

= nth term

a 1 = first term

n = post number

d = normal difference

We can find the difference by using this formula

an =a1 + (n - 1) x d

129 = 3 + (19 - 1) * d

129 = 3 + 18d

129=21xd

129/21 = d

d = 129/21

d = 6.14

Now let's find the sequence

first post

a1

Second post:

a 2 =a 1 + d

=3+6.14

=9.14

Third post:

a3 =a1 + 2d

=3+2×6.14-15.28

Fourth post:

a4 =a1 + 3d

=3+3×6.14

= 21.42

Fifth post:

a5=a1 + 4d

=3+4x6.14

= 27.56

Sixth term:

a6 =a1 +5d

=3+5×6.14

= 33.7

Thus we get all the 19 terms of this sequence can do

so the sequence is

3, 9.14, 15.28,21.42, 27.56, 33.7,39.84,

45.98,52.12, 58.26, 64.4,

70.54, 76.68, 82.82 , 88.96 ,95.1, 101.24,107.38, 113.52,119.66,129.8