Question

In an ap of 21 terms ,the sum of the first 3 terms is -33 and that of the middle 3 is 75.what is the sum of the ap ?

Answer 1

Answer:

525

Step-by-step explanation:

No of terms in the a.p=21

Sum of a1+a2+a3=-33

(a+(1-1)d)+(a+(2-1)d)+(a+(3-1)d)= - 33

a+a+d+a+2d = - 33

3a+3d = - 33

a+d = - 11 - - - - - - - (1)

Middle term =(n+1)/2

Middle term = 11th term

Middle 3 will be 10th term, 11th term and 12th term

(a+(10-1)d)+(a+(11-1)d)+(a+(12-1)d)=75

a+9d+a+10d+a+11d=75

3a+30d=75

a+10d=25---------(2)

(2)-(1)

10d-d=25-(-11)

9d=36

d=4

Substituting in (1),

a+4=-11

a=-15

21 th term = (-15+(21-1)4)

=(-15+80)

=65

Sum of ap =n/2(a+l)

Where l is last term which is 65

Sum = 21/2(-15+65)

Sum=21/2(50)

Sum=525

The sum of the a.p is 525

Answer 2

${\boxed{\boxed{\mathtt{Given}}}}$

→ Total terms in AP sequence ( n ) = 21

→ Sum of first 3 terms = - 33

→ Sum of the 3 middle terms = 75 .

${\boxed{\boxed{\mathtt{To\: Find}}}}$

→ Sum of all 21 terms of this given AP sequence .

${\boxed{\boxed{\mathtt{Solution}}}}$

Let's assume the AP sequence as

a , a + d , a+2d , + .................. a + 20 d .

So from this sequence we conclude that

⇝ First term of AP = a

⇝ Common difference = d

⇝ Last term = a + 20d

Sum of 1st 3 term = - 33

⇝ ( a )+ (a + d) + ( a + 2d ) = - 33

⇝ 3a + 3d = - 33

⇝ 3 ( a + d ) = 3 ( - 11)

a + d = -11. eq 1st

Sum of 3 middle term = 75 .

⇝ 3 Middle terms of this sequence = a+9d , a+10d , a+11d

⇝ a+9d + a+10d + a+11d = 75

⇝ 3a + 30d = 75

⇝ 3 ( a + 10d) = 3 ( 25)

a + 10 d = 25. eq 2nd

Substracting eq 1st from 2nd .

⇝ a + 10d - a - d = 25 + 11

⇝ 9d = 36

36/9 = d

d = 4

Substituting value of d in eq 1st

⇝ a + 4 = - 11

⇝ a = - 11 - 4

a = - 15

Last term of AP = a + 20d

⇝ -15 +20(4)

⇝ -15 + 80 = 65

Last term of AP ( l ) = 65

S 21 = 21/2 ( -15 + 65)

⇝ 21/2(50)

⇝ 21 × 25

⇝ 525