If Sn denotes the sum of the first n terms in an Arithmetic Progressio...
Use Sn = (n/2)[ 2a + (n-1)d] and Tn = a + (n – 1) d
S1/S4 = 1/10 = a/ (4/2) [2a + 3d]
6a = 6d or a = d
Therefore T1/T4 = a/ (a+ 3d) = a/4a = 1/4
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If Sn denotes the sum of the first n terms in an Arithmetic Progressio...
Given:
- S1:S4 = 1:10
- S(n) = n/2[2a + (n-1)d] (where a is first term, d is common difference)
- Find a : d
Solution:
1. Using the formula for sum of n terms in an AP
- S1 = a
- S4 = 4/2[2a + 3d] = 2[2a + 3d] = 4a + 6d
- S1:S4 = 1:10
- So, a:(4a + 6d) = 1:10
- Simplifying, we get: 6a = 6d
- Therefore, a:d = 1:6
2. Using the ratio of sum of terms
- S1:S4 = 1:10
- The sum of first 4 terms is: S4 = 4/2[2a + 3d] = 2[2a + 3d] = 4a + 6d
- The sum of first term is: S1 = a
- Therefore, a:S4 = S1:S4 = 1:10/S1:S4
3. Equating the two ratios
- a:S4 = 1:10/S1:S4
- a:S4 = 1:10/1:10
- a:S4 = 1:1
- Therefore, a = S4
4. Using the formula for sum of n terms in an AP
- S4 = 4/2[2a + 3d] = 2[2a + 3d] = 4a + 6d
- But, a = S4
- So, S4 = 4a + 6d = 4S4 + 6d
- Simplifying, we get: a = 6d
- Therefore, a:d = 6d:d = 6:1
5. Comparing the two values of a:d
- a:d = 1:6 (from step 1)
- a:d = 6:1 (from step 4)
- Therefore, a:d = 1:6 = 6:1
- The required ratio is a:d = 1:4.
Therefore, the answer is option C) 1:4.