Advertisements
Advertisements
Question
In a Arithmetic Progression (A.P.) the fourth and sixth terms are 8 and 14 respectively. Find that:
(i) first term
(ii) common difference
(iii) sum of the first 20 terms.
Advertisements
Solution 1
Let the first term of the sequence is a and the common difference is d.
a4 = a +3d = 8 ...(1)
a6 = a + 5d = 14 ...(2)
- - -
_______________________________
-2d = - 6
d = 3
Put d = 3 in equation (1)
a + 3 × 3 = 8
a = - 1
∴ (i) First term (a) = –1
(ii) Common difference (d) = 3
(iii) Sum of the first 20 terms = Sn `= n/2 [2a + (n-1)d]`
` = 20/2 [2 xx (-1) + 19 xx 3]`
` = 20/2 [-2+57]`
= 10 × 55 = 550
Solution 2
Let the first term of the sequence is a and the common difference is d.
a4 = a +3d = 8 ...(1)
a6 = a + 5d = 14 ...(2)
- - -
_______________________________
-2d = - 6
d = 3
Put d = 3 in equation (1)
a + 3 × 3 = 8
a = - 1
∴ (i) First term (a) = –1
(ii) Common difference (d) = 3
(iii) Sum of the first 20 terms = Sn `= n/2 [2a + (n-1)d]`
` = 20/2 [2 xx (-1) + 19 xx 3]`
` = 20/2 [-2+57]`
= 10 × 55 = 550
Solution 3
Let the first term of the sequence is a and the common difference is d.
a4 = a +3d = 8 ...(1)
a6 = a + 5d = 14 ...(2)
- - -
_______________________________
-2d = - 6
d = 3
Put d = 3 in equation (1)
a + 3 × 3 = 8
a = - 1
∴ (i) First term (a) = –1
(ii) Common difference (d) = 3
(iii) Sum of the first 20 terms = Sn `= n/2 [2a + (n-1)d]`
` = 20/2 [2 xx (-1) + 19 xx 3]`
` = 20/2 [-2+57]`
= 10 × 55 = 550
Solution 4
Let the first term of the sequence is a and the common difference is d.
a4 = a +3d = 8 ...(1)
a6 = a + 5d = 14 ...(2)
- - -
_______________________________
-2d = - 6
d = 3
Put d = 3 in equation (1)
a + 3 × 3 = 8
a = - 1
∴ (i) First term (a) = –1
(ii) Common difference (d) = 3
(iii) Sum of the first 20 terms = Sn `= n/2 [2a + (n-1)d]`
` = 20/2 [2 xx (-1) + 19 xx 3]`
` = 20/2 [-2+57]`
= 10 × 55 = 550
Solution 5
Let the first term of the sequence is a and the common difference is d.
a4 = a +3d = 8 ...(1)
a6 = a + 5d = 14 ...(2)
- - -
_______________________________
-2d = - 6
d = 3
Put d = 3 in equation (1)
a + 3 × 3 = 8
a = - 1
∴ (i) First term (a) = –1
(ii) Common difference (d) = 3
(iii) Sum of the first 20 terms = Sn `= n/2 [2a + (n-1)d]`
` = 20/2 [2 xx (-1) + 19 xx 3]`
` = 20/2 [-2+57]`
= 10 × 55 = 550
APPEARS IN
RELATED QUESTIONS
Find the sum of the following arithmetic progressions:
`(x - y)/(x + y), (3x - 2y)/(x + y), (5x - 3y)/(x + y),...` to n terms
Find the sum of the first 15 terms of the following sequences having nth term as an = 3 + 4n.
What is the 5th term form the end of the AP 2, 7, 12, ..., 47?
Find the sum of the following APs:
9, 7, 5, 3, ... to 14 terms.
Write an A.P. whose first term is a and the common difference is d in the following.
a = 10, d = 5
Choose the correct alternative answer for the following question .
In an A.P. first two terms are –3, 4 then 21st term is ...
Two A.P.’s are given 9, 7, 5, ... and 24, 21, 18, ... If nth term of both the progressions are equal then find the value of n and nth term.
A piece of equipment cost a certain factory Rs 600,000. If it depreciates in value, 15% the first, 13.5% the next year, 12% the third year, and so on. What will be its value at the end of 10 years, all percentages applying to the original cost?
Yasmeen saves Rs 32 during the first month, Rs 36 in the second month and Rs 40 in the third month. If she continues to save in this manner, in how many months will she save Rs 2000?
Which term of the Arithmetic Progression (A.P.) 15, 30, 45, 60...... is 300?
Hence find the sum of all the terms of the Arithmetic Progression (A.P.)
