Question

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

1. A series with six terms, the first term is five, and the common difference is five.

2. A series with six terms, the first term is nine and the common difference is twelve.

3. A series with five terms, the first term is 5.7 and the common difference is 1.4.

4. Find the sum of the first 21 terms in a series where a = 20 and t21= 400.

5. Find the sum of the first 100 terms in a series where a = 0 and t100= 99.

6. Find the sum of the first 50 terms in a series where a = 4 and t50= 196.

7. Find the sum of the first 100 terms in a series where a = 0 and t100= 99.

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

Step-by-step explanation:

I hope it was helpful to you

Answer 2

The answer are 1) 105,  2) 234, 3) 42.5,  4) 4410, 5) 4950, 6) 5000,

7) 4950, 8)140, 9)220, 10) 2325.

Given: 1) In AP, Number of terms = 6, a = 5 and d = 5

2) In AP, Number of terms = 6, a = 9 and d = 12

3) In AP, Number of terms = 5, a = 5.7 and d = 1.4

4) In AP, = 20, t21 = 400  

5) In AP,a = 0 and t₁₀₀ = 99  

6) In AP, a = 4 and t₅₀= 196  

7) In AP, a = 0 and t₁₀₀= 99

8) In AP, t₇ = 7 and t₁₀ = 13

9) In AP, t₅ = 15 and t₁₀ = 45

10) Series of first 30 positive multiples of 5.

To find: sum of terms of Arithmetic series  

As we know sum of n terms in AP Sn= n/2[2a+(n-1)d]

1)  Number of terms = 6  where a = 5 and d = 5

$S_{6} = \frac{6}{2} [2(5)+(6-1)5]= 3[10+5(5)] = 3[10+25] = 3(35) =105$

2) Number of terms =6 where a = 9 and d = 12.

$S_{6} = \frac{6}{2} [2(9)+(6-1)12]= 3[18+5(12)] = 3[18+60] = 3(78) =234$

3) Number of terms = 5 where a = 5.7 and d = 1.4

$S_{5} =\frac{5}{2} [2(5.7)+(5-1)1.4]= 2.5[11.4+4(1.4)]= 2.5[11.4+5.6] = 2.5(17) =42.5$

4) a = 20 and  t₂₁ = 400 ,

t₂₁ = 20 + (21 - 1)d = 400

⇒ 20 +19d = 400  ⇒ 19d = 380

Therefore, Sum of 21 terms =  

$S_{21} = \frac{21}{2} [2(20)+(20-1)d]= 10.5[40+19d] =10.5[40+380] =10.5(420) =4410$

5) In series  a = 0 and t₁₀₀ =  99  

⇒ t₁₀₀ = 0+(100-1)d = 99  ⇒ 99d = 99  

Therefore, Sum of 100 terms =  

$S_{100} = \frac{100}{2} [2(0)+(100-1)d]= 50[0+99d] =50[99] =4950$

6) In series a = 4 and t₅₀ = 196,

⇒ T₅₀ = 4+(50-1)d = 196    ⇒  49d = 192

Therefore,  Sum of 50 terms =  

$S_{50} = \frac{50}{2} [2(4)+(50-1)d]= 25[8+49d] =25[8+192] =25(200) =5000$

7) In the series,  a = 0 and t₁₀₀= 99

⇒ t₁₀₀ = 0+(100-1)d = 99  ⇒ 99d = 99  

Therefore, Sum of 100 terms =  

$S_{100} = \frac{100}{2} [2(0)+(100-1)d]= 50[0+99d] =50[99] =4950$

8) In a series, t₇ = 7 and t₁₀ = 13  

⇒  t₇= a+(7-1)d = 7                 ⇒   T₁₀ = a+(10-1)d = 13

⇒ a+6d = 7 _(1)                     ⇒     a + 9d = 13 _(2)  

Subtract (1) from (2) ⇒ a + 12d (a+6d) = 13 - 7  

⇒ a +6d - a - 9d = - 6

⇒   - 3d = -6     ⇒ d = 2          

Substitute d = 1 in (1)  

⇒  a+6(2) = 7

⇒  a + 12 = 7     ⇒ a = 5

Therefore, In given series a= 5 and d = 2  

Sum of first 10 numbers =

$S_{10} = \frac{10}{2} [2(5)+(10-1)2]= 5[10+9(2)] =5[10+18] =5(28) = 140$  

9) In the series t₅ = 15 and t₁₀ = 45

⇒  t₅ = a+(5-1)d = 15                 ⇒   T₁₀ = a+(10-1)d = 13

⇒  a+4d = 15 _(1)                        ⇒    a + 9d = 45_(2)  

Subtract (1) from (2)  ⇒ a + 9d - (a+4d) = 45 - 15

⇒ a + 9d - a - 4d = 30

⇒ 5d = 30  ⇒ d = 6  

Substitute d = 6 in (1)

⇒ a + 4(6)  = 15

⇒ a + 20 = 15   ⇒ a = -5

Therefore, In given series a = - 5 and d = 6  

Sum of 10 terms  

$S_{10} = \frac{10}{2} [2(-5)+(10-1)6]= 5[-10+9(6)] =5[-10+54] =5(44) =220$

10) Multiples of 5  =  5, 10, 15, 20 ..  is a AP (∵ common difference is 5 )

Therefore,   a = 5 and d = 5

$S_{30} = \frac{30}{2} [2(5)+(30-1)5]= 15[10+29(5)] =15[10+145] =15(155) =2325$

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