Question

HELP !!! ASAP HELP ME OUT !

Directions: Find the sum for each of the following arithmetic series below. Use the summation formula to find the sums.

use the formulas:
Sn=(n/2) (a + tn) , tn = a + (n -1)d tn

8. Find the sum of the first 10 terms in an arithmetic series where t7= 7 and t10= 13.

9. Find the sum of the first 10 terms in an arithmetic series where t5= 15 and t10= 45.

10. Find the sum of the first 30 positive multiples of 5.

Answer 1

Answer:

$8^{th} Answer = 40\\\\ 9^{th} Answer = 180\\ \\ 10^{th} Answer = 2325$

Step-by-step explanation:

8. We know,
     $T_{n} = a + (n-1)d$
     $T_{7} = a + (7-1)d$
So, $7 = a + 6d$

Similarly with $T_1_0$ ;
$T_{10} = a + (10-1)d$

So, $13 = a + 9d$

Now, by Subtracting these two we will get :
$6= 3d$

$d = 2$

Now put this d in any equation let's say 1st one so;
$T_7 = a + 6 * 2$

7 = a + 12

a = -5
Now, We know, Sn=(n/2) (a + $T_n$)

So,
$S_1_0 = (10/2) * (a + T_1_0 )$

$S_1_0 = 5 * ( -5 + 13 )$

∴ $S_1_0 = 40$ = Answer

9. We know,
     $T_{n} = a + (n-1)d$
     $T_{5} = a + (5-1)d$
So, $15 = a + 4d$

Similarly with $T_1_0$ ;
$T_{10} = a + (10-1)d$

So, $45 = a + 9d$

Now, by Subtracting these two we will get :
$30 = 5d$

$d = 6$

Now put this d in any equation let's say 1st one so;
$T_5 = a + 4 * 6$

15 = a + 24

a = -9
Now, We know, Sn=(n/2) (a + $T_n$)

So,
$S_1_0 = (10/2) * (a + T_1_0 )$

$S_1_0 = 5 * ( -9 + 45 )$

∴ $S_1_0 = 180$ = Answer

10. Here, a = 5

     d = 5
     n = 30

So We know, Sn=(n/2) (a + $T_n$)

$S_3_0 = (30/2) ( a + a + (n-1) d )$

$S_3_0 = 15 * (10 + 29*5)\\ S_3_0 = 15 * (155)\\S_3_0 = 2325$

∴ $S_3_0 = 2325$ = Answer