Let the first instalment be ( a ) rupees.
Common difference ( d = 10 ).
Number of instalments ( n = 40 ).
Total sum ( S_{40} = 10400 ).
[ S_{40} = \frac{40}{2}\bigl[2a + (40-1)\cdot 10\bigr] = 20\bigl[2a + 390\bigr] = 40a + 7800. ]
[ 40a + 7800 = 10400 \implies 40a = 2600 \implies a = 65. ]
Answer: Rs. 65