solve:-
Term at 6 position is 10 *5^(6-1)
= 10 *5⁽⁵⁾
= 10 * 3125
= 31250

Now,
The sum of 6 to 10 position is 31250*[1 - 5⁽⁵⁾] /[1-5]
= (31250)[ 1 - 5⁽⁵⁾]/[1-5]

= (31250)(-3124)/(-4)

= (31250)(781)

ie. S₆ to S₁₀ = 24406250

Second method:-
Term at 6 position is 10 *5^(6-1)
= 10 *5⁽⁵⁾

= 10 * 3125
= 31250 (=t1)

Term at 7 position is 10 *5^(7-1)
= 10 *5⁽⁶⁾

= 10 * 15625
= 156250 (=t2)

Term at 8 position is 10 *5^(8-1)
= 10 *5⁽⁷⁾

= 10 * 78125
= 781250 (=t3)

Term at 9 position is 10 *5^(9-1)
= 10 *5⁽⁸⁾

= 10 * 390625
= 3906250 (=t4)

Term at 10 position is 10 *5^(10-1)
= 10 *5⁽⁹⁾

= 10 * 1953125
= 19531250 (=t5)

so the sequance of numbers from position 6 to 10 are 31250, 156250, 781250, 3906250, 19531250
= 31250 + 156250 + 781250 + 3906250 + 19531250;

= 24406250