Question

The 104th term and 4th term of an A.P. are 125 and 0. Find the sum of first 35 terms.
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Answer 1

Answer:

hi.

Step-by-step explanation:

given a 104= 125

a4=0

now.

to find- S35.

a + 103d=125

a+3d = 0

subtracting we get.

100d =125

d=125/100=5/4

now. d=5/4

a= -15/4..

S35= 35/2(2×-15/4 +34×5/4)

=35/2(-15/2+85/2)

= 35/2×35

=1225/2=612.5

therefore sum of first 35 terms is 612.5

Answer 2

Answer:

$\underline{\Huge{\texttt{612.5}}}$

Step-by-step explanation:

$\text{104}^{\text{th}} \text{ term = a + 103d = 125} \text{  ....(1)}\\\text{4}^{\text{th}} \text{ term = a + 3d = 0} \text{  ....(2)}$

Subtracting eqⁿ (1) & (2), we get:

$\text{100d = 125}\\\implies \text{d = }\frac{125}{100} \text{ = 1.25}$

$\therefore \text{ a + 3d = 0}\\\implies \text{a = -3d} \implies \text{a = -3} \cdot \text{(1.25)} \implies \text{a = \underline{\underline{-3.75}}}$

$\text{S}_\text{n} = \frac{n}{2} (2a + (n - 1)d)$

$S_{35} = \frac{35}{2} (2\cdot(-3.75) + 34 \cdot (1.25))\\\\= \frac{35}{2} ((-7.5) + (42.5))\\\\= \frac{35}{2}\cdot (35)\\\\\implies 17.5 \times 35 = 612.5$