Question

The sum of the first 11 terms of an A. P is 19 and the sum of first 19 terms is 11. Find the sum of the first 30 terms.​

Answer 1

Step-by-step explanation:

Given :-

The sum of the first 11 terms of an A. P is 19 and the sum of first 19 terms is 11

To Find :-

Sum of 30 terms

Solution :-

We know that

S_n = n/2[2a + (n - 1)d]

For first 11 terms

19 = 11/2[2a + (11 - 1)d]

19 = 5.5[2a + 10d]

19 = 11a + 55d (1)

For first 19 terms

11 = 19/2[2a + (19 - 1)d]

11 = 9.5[2a + 18d]

11 = 19a + 171d (2)

On subtracting

19 - 11 = 11a + 55d - 19a - 171d

8 = -8a - 116d

1 = -a - 14.5d

a + 14.5d = -1

a = -14.5d

Now,

Sum of 30 terms

30/2[2(-14.5d) + (30 - 1)d]

15[-29d + 29d]

15[0]

0

Answer 2

Answer:

The sum of first $30$ terms of the AP is $-435$

Step-by-step explanation:

We have the equation for the sum of first n terms of an AP, is

$S_{n} =\frac{n}{2} [2a+(n-1)d]$

Therefore sum of first eleven terms of an AP, is

$S_{11} =\frac{11}{2} [2a+10d]=19$

That is $11a+5d=19$ . . . .($1$)(cancelling throughout by the denominator two.)

Also given the sum of first $19$ terms is $11$.

That is,

$S_{19} =\frac{19}{2} [2a+18d]$

     $=19a+9d=11$ . . . . $(2)$

Now we can find the values of $a$   and   $d$  from the equations $(1),(2)$, and then write the equation for sum of first thirty terms and substitute the values of $a$  and   $d$.

Solving equations $(1),(2)$ by the method elimination. So, multiply equation $(1)$ with $9$  and $(2)$ with $5$ to make the coefficient of $d$ equal in both the equations and hence we can eliminate $d$.

subtract $(2)*5$ from $(1)*9$

$(1)*9$⇒

             $99a+45d=171$

$(2)*5$⇒

              $95a+45d=55$

$(1)*9-(2)*5$⇒

                $4a=171-55=116\\\\a=116/4=29$

applying the value of $a$ in $(1)$,  

                $11*29+10d=19\\\\10d=19-319\\\\d=-300/10=-3$

Now consider sum of $30$ terms

$S_{30} =\frac{30}{2} [2a+29d]$,          we get  $a=29$  and  $d=-3$, therefore

     $=15[29*2+29*-3]\\\\=15[-29]\\\\=-435$

Hence the answer

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