Question

Find the sum of all natural number from 50 to 250 which are divisible
by 6 and Find t tu
Find A.P.
Find n
Find S
33
Find t
13​

Answer 1

Answer:

A.P. is 54, 60, 66 ...............246

d= 6= common difference,. n= no.of terms

a= first term= 54 , last term= l =246

Step-by-step explanation:

since, l = a+ ( n-1)d

246= 54 +( n-1)6

(n-1)6= 192

n-1 =192/6 =32

n=32+1= 33

n = 33

s = n/2(a+l)

s=( 33/2)*(54+246) = 33/2*(300) =33*150= 4950

s = 4950

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Answer 2

$\huge{\green{\underline{\underline{Answer:-}}}}$

Proper question:

Find the sum of all natural number from 50 to 250 which are divisible by 6.Find t13?

Given:

Natural numbers from 50 to 250

To find:

Sum of all natural numbers from 50 to 250 which are divisible by 6 = ?

t13 = ?

Solution:

Numbers divisible by 6 from 50 to 250 are

54, 60 , 66, 72 , ............., 246

General formula for tn = a + (n - 1) d

a = first term

d = common difference between terms

Here, a = t1 = 54

tn = 246

d = t2 - t1

= 66 - 60

= 6

tn = = a + (n - 1) d

246 = 54 + (n - 1) 6

246 = 54 + 6n - 6

246 = 48 + 6n

246 - 48 = 6n

198 = 6n

$\\ \\ n = \dfrac{198}{6} \\ \\ n = 33$

Sn =n/2 (2a + (n - 1) d

S33 = 33/2 (2 × 54 + (33 - 1) 6

S33 = 33/2 (108 + 32 × 6)

S33 = 33/2 ( 108 + 92)

S33 = 33/2 × 300

S33 = 4950

t13 = a + (n - 1) d

t13 = 54 + (13 - 1) 6

t13 = 54 + 12 × 6

t13 = 54 + 72

t13 = 126

Answer:

Thus, sum of all natural numbers from 50 to 250 is 4950 and t13 is 126

Basic point:

☆ Sum of terms formula,

--> Sn = n/2 (2a + (n-1) d

☆ General formula of terms,

--> tn = a + (n -1 ) d

☆ Try to solve more such questions to get good hold on it